Question
When Andy plays tennis, \(65\% \) of his first serves go into the correct area of the court.
If the first serve goes into the correct area, his chance of winning the point is \(90\% \).
If his first serve does not go into the correct area, Andy is allowed a second serve and, of these, \(80\% \) go into the correct area.
If the second serve goes into the correct area, his chance of winning the point is \(60\% \).
If neither serve goes into the correct area, Andy loses the point.
Complete the tree diagram below.
[2]
Find the probability that Andy loses the point.[4]
Answer/Explanation
Markscheme
(A1)(A1) (C2)[2 marks]
\(0.65 \times 0.1\) (\( = 0.065\)) (A1)
\(0.35 \times 0.8 \times 0.4\) (\( = 0.112\)) (A1)
\(0.35 \times 0.2 \times 1\) the 1 can be implied (\( = 0.07\)) (A1)(ft)
0.247 (A1)(ft) (C4)
Note: No (ft) for any probabilities greater than 1.[4 marks]
Question
The grades obtained by a group of \(20\) IB students are listed below:
Complete the following table for the grades obtained by the students.
[2]
Write down the modal grade obtained by the students.[1]
Calculate the median grade obtained by the students.[2]
One student is chosen at random from the group.
Find the probability that this student obtained either grade \(4\) or grade \(5\).[1]
Answer/Explanation
Markscheme
(A2) (C2)
Notes: Award (A1) for three correct. Award (A0) for two or fewer correct.[2 marks]
\({\text{Mode}} = 6\) (A1)(ft) (C1)[1 mark]
\({\text{Median}} = 4.5\) (M1)(A1)(ft) (C2)
Note: (M1) for attempt to order raw data (if frequency table not used) or (M1) halfway between 10th and 11th result.[2 marks]
\(\frac{7}{{20}}{\text{ }}(0.35{\text{, }}35\% )\) (A1)(ft) (C1)[1 mark]
Question
A fair six-sided die has the numbers 1, 2, 3, 4, 5, 6 written on its faces. A fair four-sided die has the numbers 1, 2, 3, and 4 written on its faces. The two dice are rolled.
The following diagram shows the possible outcomes.
Find the probability that the two dice show the same number.[2]
Find the probability that the difference between the two numbers shown on the dice is 1.[2]
Find the probability that the number shown on the four-sided die is greater than the number shown on the six-sided die, given that the difference between the two numbers is 1.[2]
Answer/Explanation
Markscheme
\(\frac{4}{{24}}\) \(\left( {\frac{1}{6},0.167,16.7{\text{ }}\% } \right)\) (A1)(A1) (C2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\(\frac{{7}}{{24}}\) \((0.292,29.2{\text{ }}\% )\) (A1)(A1)(ft) (C2)
Note: Award (A1)(ft) from the denominator used in (a).[2 marks]
\(\frac{{3}}{{7}}\) \((0.429,42.9{\text{ }}\% )\) (A1)(A1)(ft) (C2)
Note: Award (A1) for numerator (A1)(ft) for denominator, (ft) from their numerator in (b).[2 marks]
Question
For events A and B, the probabilities are \({\text{P}}(A) = \frac{4}{13}\) and \({\text{P}}(B) = \frac{5}{13}\).
If events A and B are mutually exclusive, write down the value of \({\text{P}} (A\cap B)\).[1]
If events A and B are independent, find the value of \({\text{P}} (A\cap B)\).[2]
If \({\text{P}} (A \cup B) = \frac{7}{13}\), find the value of \({\text{P}} (A \cap B)\).[3]
Answer/Explanation
Markscheme
\({\rm{P}}(A \cap B) = 0\) (A1) (C1)[1 mark]
\({\rm{P}}(A \cap B) = {\rm{P}}(A) \times {\rm{P}}(B)\)
\( = \frac{4}{{13}} \times \frac{5}{{13}}\) (M1)
Note: Award (M1) for product of two fractions, decimals or percentages.
\({\rm{P}}(A \cap B) = \frac{{20}}{{169}} (= 0.118)\) (A1) (C2)[2 marks]
\(\frac{7}{{13}} = \frac{4}{{13}} + \frac{5}{{13}} – {\rm{P}}(A \cap B)\) (M1)(M1)
Notes: Award (M1) for \(\frac{4}{{13}} + \frac{5}{{13}}\) seen, (M1) for subtraction of \(\frac{7}{{13}}\) shown.
OR
Award (M1) for Venn diagram with 2 intersecting circles, (A1) for correct probabilities in diagram.
\({\rm{P}}(A \cap B) = \frac{2}{{13}}( = 0.154)\) (A1) (C3)[3 marks]
Question
A weighted die has 2 red faces, 3 green faces and 1 black face. When the die is thrown, the black face is three times as likely to appear on top as one of the other five faces. The other five faces have equal probability of appearing on top.
The following table gives the probabilities.
Find the value of
(i) m;
(ii) n.[2]
The die is thrown once.
Given that the face on top is not red, find the probability that it is black.[2]
The die is now thrown twice.
Calculate the probability that black appears on top both times.[2]
Answer/Explanation
Markscheme
(i) m = 1 (A1)
(ii) n = 3 (A1) (C2)
Note: Award (A0)(A1)(ft) for \(m = \frac{1}{8}, n = \frac{3}{8}\).
Award (A0)(A1)(ft) for m = 3, n = 1.[2 marks]
\({\rm{P}}(B/R’) = \frac{{\frac{3}{8}}}{{\frac{6}{8}}} = \frac{3}{6}\left( {\frac{1}{2},50\% ,0.5} \right)\) (M1)(A1)(ft) (C2)
Note: Award (M1) for correctly substituted conditional probability formula or for 6 seen as part of denominator.[2 marks]
\({\rm{P}}(B,B) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{{64}}(0.141)\) (M1)(A1)(ft) (C2)
Note: Award (M1) for product of two correct fractions, decimals or percentages.
(ft) from their answer to part (a) (ii).[2 marks]
Question
The probability that it will snow tomorrow is 0.3.
If it snows tomorrow the probability that Chuck will be late for school is 0.8.
If it does not snow tomorrow the probability that Chuck will be late for school is 0.1.
Complete the tree diagram below.
[3]
Find the probability that it does not snow tomorrow and Chuck is late for school.[1]
Find the probability that Chuck is late for school.[2]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair.[3 marks]
\(0.7 \times 0.1\)
\( = 0.07(\frac{7}{{100}},{\text{ }}7\% )\) (A1)(ft) (C1)[1 mark]
\(0.3 \times 0.8 + 0.07\) (M1)
\( = 0.31(\frac{{31}}{{100}},{\text{ }}31\% )\) (A1)(ft)
Note: In (b) and (c) follow through from sensible answers only i.e. not a probability greater than one. (C2)[2 marks]
Question
A class consists of students studying Spanish or French or both. Fifteen students study Spanish and twelve study French.
The probability that a student studies French given that she studies Spanish is \(\frac{{7}}{{15}}\).
Draw a Venn diagram in the space below to illustrate this information.[3]
Find the probability that a student studies Spanish given that she studies one language only.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for a labeled Venn diagram with appropriate sets.
(A1) for 7, (A1) for 8 and 5.[3 marks]
\({\text{P (Spanish / one language only)}} = \frac{{\frac{8}{{20}}}}{{\frac{8}{{20}} + \frac{5}{{20}}}}\) (M1)(A1)(ft)
Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution. Follow through from their Venn diagram.
\( = \frac{8}{{13}}(0.615,{\text{ }}61.5\% )\) (A1)(ft)
OR
\({\text{P}}{\text{ (Spanish / one language only)}} = \frac{8}{{8 + 5}}\) (A1)(ft)(M1)
Note: Award (A1) for their correct numerator, (M1) for correct recognition of regions. Follow through from their Venn diagram.
\( = \frac{8}{{13}}(0.615,{\text{ }}61.5\% )\) (A1)(ft) (C3)[3 marks]
Question
Maria travels to school either by walking or by bicycle. The probability she cycles to school is 0.75.
If she walks, the probability that she is late for school is 0.1.
If she cycles, the probability that she is late for school is 0.05.
Complete the tree diagram below, showing the appropriate probabilities.
[3]
Find the probability that Maria is late for school.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for 0.25, (A1) for 0.1 and 0.9, (A1) for 0.05 and 0.95[3 marks]
\({\text{P}}({\text{late}}) = 0.25 \times 0.1 + 0.75 \times 0.05\) (A1)(ft)(M1)
Note: Award (A1)(ft) for two correct products from their diagram and award (M1) for addition of their two products.
\( = 0.0625\left( {\frac{1}{{16}},{\text{ }}6.25\% } \right)\) (A1)(ft) (C3)[3 marks]
Question
In a research project on the relation between the gender of 150 science students at college and their degree subject, the following set of data is collected.
Find the probability that a student chosen at random is male.[2]
Find the probability that a student chosen at random is either male or studies Chemistry.[2]
Find the probability that a student chosen at random studies Physics, given that the student is male.[2]
Answer/Explanation
Markscheme
\( = \frac{{91}}{{150}}(0.607,{\text{ }}60.6\,\% ,{\text{ }}60.7\,\% )\) (A1)(A1) (C2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\( = \frac{{111}}{{150}}\left( {\frac{{37}}{{50}},{\text{ }}0.74,{\text{ }}74\,\% } \right)\) (A1)(ft)(A1) (C2)
Note: Award (A1)(ft) for their numerator in (a) +20 provided the final answer is not greater than 1. (A1) for denominator.[2 marks]
\(\frac{{16}}{{91}}(0.176,{\text{ }}17.6\,\% )\) (A1)(A1)(ft) (C2)
Note: Award (A1) for numerator and (A1)(ft) for denominator. Follow through from their numerator in (a) provided answer is not greater than 1.[2 marks]
Question
Let \({\text{P}}(A) = 0.5\), \({\text{P}}(B) = 0.6\) and \({\text{P}}(A \cup B) = 0.8\).
Find \({\text{P}}(A \cap B)\).[2]
Find \({\text{P}}(A|B)\).[2]
Decide whether A and B are independent events. Give a reason for your answer.[2]
Answer/Explanation
Markscheme
\(0.8 = 0.5 + 0.6 – {\text{P}}(A \cap B)\) (M1)
\({\text{P}}(A \cap B) = 0.3\) (A1) (C2)
Note: Award (M1) for correct substitution, (A1) for correct answer.[2 marks]
\({\text{P}}(A|B) = \frac{{0.3}}{{0.6}}\) (M1)
= 0.5 (A1)(ft) (C2)
Note: Award (M1) for correct substitution in conditional probability formula. Follow through from their answer to part (a), provided probability is not greater than one.[2 marks]
\({\text{P}}(A \cap B) = {\text{P}}(A) \times {\text{P}}(B)\) or 0.3 = 0.5 × 0.6 (R1)
OR
\({\text{P}}(A|B) = {\text{P}}(A)\) (R1)
they are independent. (Yes) (A1)(ft) (C2)
Note: Follow through from their answers to parts (a) or (b).
Do not award (R0)(A1).[2 marks]
Question
A survey was carried out in a group of 200 people. They were asked whether they smoke or not. The collected information was organized in the following table.
One person from this group is chosen at random.
Write down the probability that this person is a smoker.[2]
Write down the probability that this person is male given that they are a smoker.[2]
Find the probability that this person is a smoker or is male.[2]
Answer/Explanation
Markscheme
\(\frac{{90}}{{200}}(0.45,{\text{ }}45{\text{ }}\% )\) (A1)(A1) (C2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\(\frac{{60}}{{90}}(0.\bar 6,{\text{ }}0.667,{\text{ }}66.\bar 6{\text{ }}\% ,{\text{ }}66.6 \ldots {\text{ }}\% ,{\text{ }}66.7{\text{ }}\% )\) (A1)(A1)(ft) (C2)
Notes: Award (A1) for numerator, (A1)(ft) for denominator, follow through from their numerator in part (a). Last mark is lost if answer is not a probability.[2 marks]
\(\frac{{90}}{{200}} + \frac{{100}}{{200}} – \frac{{60}}{{200}}\) (M1)
Note: Award (M1) for correct substitution in the combined events formula. Follow through from their answer to part (a).
\( = \frac{{130}}{{200}}(0.65,{\text{ }}65{\text{ }}\% )\) (A1)(ft)
OR
\(\frac{{60}}{{200}} + \frac{{40}}{{200}} + \frac{{30}}{{200}}\) (M1)
Note: Award (M1) for adding the correct fractions.
\( = \frac{{130}}{{200}}(0.65,{\text{ }}65{\text{ }}\% )\) (A1)
OR
\(1 – \frac{{70}}{{200}}\) (M1)
Note: Award (M1) for subtraction of correct fraction from 1.
\( = \frac{{130}}{{200}}(0.65,{\text{ }}65{\text{ }}\% )\) (A1) (C2)
[2 marks]
Question
The probability that it rains today is \(0.4\) . If it rains today, the probability that it will rain tomorrow is \(0.8\) . If it does not rain today, the probability that it will rain tomorrow is \(0.7\) .
Complete the tree diagram below.
[3]
Calculate the probability of rain tomorrow.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair.[3 marks]
\(0.4 \times 0.8 + 0.6 \times 0.7\) (A1)(ft)(M1)
Notes: Award (A1)(ft) for two consistent products from tree diagram, (M1) for addition of their products. Follow through from their tree diagram provided all probabilities are between 0 and 1.
\( = 0.74\) (A1)(ft) (C3)[3 marks]
Question
A survey was carried out at an international airport. A number of travellers were interviewed and asked for their flight destinations. The results are shown in the table below.
One traveller is to be chosen at random from all those interviewed.
Find the probability that this traveller was going to Africa.[2]
One female traveller is to be chosen at random from all those interviewed.
Find the probability that this female traveller was going to Asia.[2]
One traveller is to be chosen at random from those not going to America.
Find the probability that the chosen traveller is female.[2]
Answer/Explanation
Markscheme
\(\frac{{108}}{{250}}{\text{ }}\left( {\frac{{54}}{{125}}{\text{, }}0.432{\text{, }}43.2\% } \right)\) (A1)(A1) (C2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\(\frac{{25}}{{106}}{\text{ }}\left( {0.236{\text{, }}23.6\% } \right)\) (A1)(A1) (C2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
\(\frac{{71}}{{170}}{\text{ }}\left( {0.418{\text{, }}41.8\% } \right)\) (A1)(A1) (C2)
Note: Award (A1) for numerator, (A1) for denominator.[2 marks]
Question
A bag contains 7 red discs and 4 blue discs. Ju Shen chooses a disc at random from the bag and removes it. Ramón then chooses a disc from those left in the bag.
Write down the probability that
(i) Ju Shen chooses a red disc from the bag;
(ii) Ramón chooses a blue disc from the bag, given that Ju Shen has chosen a red disc;
(iii) Ju Shen chooses a red disc and Ramón chooses a blue disc from the bag.[3]
Find the probability that Ju Shen and Ramón choose different coloured discs from the bag.[3]
Answer/Explanation
Markscheme
(i) \(\frac{7}{{11}}\) (\(0.636\), \(63.6\% \)) (\(0.636363 \ldots \)) (A1) (C1)
(ii) \(\frac{4}{{10}}\) \(\left( {\frac{2}{5}{\text{, }}0.4{\text{, }}40\% } \right)\) (A1) (C1)
(iii) \(\frac{{28}}{{110}}\) \(\left( {\frac{{14}}{{55}}{\text{, }}0.255{\text{, }}25.5\% } \right)\) \(0.254545 \ldots \) (A1)(ft) (C1)
Note: Follow through from the product of their answers to parts (a) (i) and (ii).[3 marks]
\(\frac{{28}}{{110}} + \left( {\frac{4}{{11}} \times \frac{7}{{10}}} \right)\) OR \(2 \times \frac{{28}}{{110}}\) (M1)(M1)
Notes: Award (M1) for using their \(\frac{{28}}{{110}}\) as part of a combined probability expression. (M1) for either adding \({\frac{4}{{11}} \times \frac{7}{{10}}}\) or for multiplying by 2.
\( = \frac{{56}}{{110}}\) \(\left( {\frac{{28}}{{55}}{\text{, }}0.509{\text{, }}50.9\% } \right)\) (\(0.509090 \ldots \)) (A1)(ft) (C3)
Note: Follow through applies from their answer to part (a) (iii) and only when their answer is between 0 and 1.[3 marks]
Question
Merryn plans to travel to a concert tomorrow. Due to bad weather, there is a 60 % chance that all flights will be cancelled tomorrow. If the flights are cancelled Merryn will travel by car.
If she travels by plane the probability that she will be late for the concert is 10 %.
If she travels by car, the probability that she will not be late for the concert is 25 %.
Complete the tree diagram below.
[1]
Find the probability that Merryn will not be late for the concert.[3]
Merryn was not late for the concert the next day.
Given that, find the probability that she travelled to the concert by car.[2]
Answer/Explanation
Markscheme
(A1) (C1)
Note: Award (A1) for 0.9 and 0.75.[1 mark]
0.4 × 0.9 + 0.6 × 0.25 (M1)(M1)
Note: Award (M1) for their two relevant products, (M1) for adding their two products.
\(0.51\left( {\frac{{51}}{{100}},{\text{ }}51\% } \right)\) (A1)(ft) (C3)
Note: Follow through from their answers to part (a).[3 marks]
\(\frac{{0.6 \times 0.25}}{{0.51}}\) (M1)
Note: Award (M1) for correctly substituted conditional probability formula.
\(0.294\left( {\frac{5}{{17}}{\text{, }}0.294117…} \right)\) (A1)(ft) (C2)
Note: Follow through from their tree diagram and their part (b).[2 marks]
Question
The probability that Tanay eats lunch in the school cafeteria is \(\frac{3}{5}\).
If he eats lunch in the school cafeteria, the probability that he has a sandwich is \(\frac{3}{{10}}\).
If he does not eat lunch in the school cafeteria the probability that he has a sandwich is \(\frac{9}{{10}}\).
Complete the tree diagram below.
[3]
Find the probability that Tanay has a sandwich for his lunch.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair of branches.
\(\frac{3}{5} \times \frac{3}{{10}} + \frac{2}{5} \times \frac{9}{{10}}\) (A1)(ft)(M1)
Notes: Award (A1)(ft) for their two correct products, (M1) for addition of their products. Follow through from their tree diagram.
\( = \frac{{27}}{{50}}(0.54,54\% )\) (A1)(ft) (C3)
Question
Alan’s laundry basket contains two green, three red and seven black socks. He selects one sock from the laundry basket at random.
Write down the probability that the sock is red.[1]
Alan returns the sock to the laundry basket and selects two socks at random.
Find the probability that the first sock he selects is green and the second sock is black.[2]
Alan returns the socks to the laundry basket and again selects two socks at random.
Find the probability that he selects two socks of the same colour.[3]
Answer/Explanation
Markscheme
\(\frac{3}{{12}}\left( {\frac{1}{4},0.25,25\% } \right)\) (A1) (C1)
\(\left( {\frac{2}{{12}}} \right) \times \left( {\frac{7}{{11}}} \right)\) (M1)
Note: Award (M1) for correct product.
\( = \frac{{14}}{{132}}\left( {\frac{7}{{66}},0.10606…,10.6\% } \right)\) (A1) (C2)
\(\left( {\frac{2}{{12}} \times \frac{1}{{11}}} \right) + \left( {\frac{3}{{12}} \times \frac{2}{{11}}} \right) + \left( {\frac{7}{{12}} \times \frac{6}{{11}}} \right)\) (M1)(M1)
Note: Award (M1) for addition of their 3 products, (M1) for 3 correct products.
\( = \frac{{50}}{{132}}\left( {\frac{25}{{66}},0.37878…,37.9\% } \right)\) (A1) (C3)
Question
Ramzi travels to work each day, either by bus or by train. The probability that he travels by bus is \(\frac{3}{5}\). If he travels by bus, the probability that he buys a magazine is \(\frac{2}{3}\). If he travels by train, the probability that he buys a magazine is \(\frac{3}{4}\).
Complete the tree diagram.
[3]
Find the probability that Ramzi buys a magazine when he travels to work.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair of branches.[3 marks]
\(\frac{3}{5} \times \frac{2}{3} + \frac{2}{5} \times \frac{3}{4}\) (A1)(ft)(M1)
Notes: Award (A1)(ft) for two consistent products from tree diagram, (M1) for addition of their products.
Follow through from their tree diagram provided all probabilities are between \(0\) and \(1\).
\(\frac{7}{{10}}{\text{ }}\left( {{\text{0.7, 70% , }}\frac{{42}}{{60}}} \right)\) (A1)(ft) (C3)[3 marks]
Question
The probability that it snows today is 0.2. If it does snow today, the probability that it will snow tomorrow is 0.6. If it does not snow today, the probability that it will not snow tomorrow is 0.9.
Using the information given, complete the following tree diagram.
[3]
Calculate the probability that it will snow tomorrow.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair of probabilities.[3 marks]
\(0.2 \times 0.6 + 0.8 \times 0.1\) (A1)(ft)(M1)
Note: Award (A1)(ft) for two correct products of probabilities taken from their diagram, (M1) for the addition of their products.
\( = 0.2{\text{ }}\left( {\frac{1}{5},{\text{ 20% }}} \right)\) (A1)(ft) (C3)
Note: Accept any equivalent correct fraction.
Follow through from their tree diagram.[3 marks]
Question
Aleph has an unbiased cubical (six faced) die on which are written the numbers
1 , 2 , 3 , 4 , 5 and 6.
Beth has an unbiased tetrahedral (four faced) die on which are written the numbers
2 , 3 , 5 and 7.
Complete the Venn diagram with the numbers written on Aleph’s die (\(A\)) and Beth’s die (\(B\)).
[2]
Find \(n(B \cap A’)\).[2]
Aleph and Beth are each going to roll their die once only. Shin says the probability that each die will show the same number is \(\frac{1}{8}\).
Determine whether Shin is correct. Give a reason.[2]
Answer/Explanation
Markscheme
(A1)(A1) (C2)
Note: Award (A1) for 2, 3, 5 in intersection, (A1) for 1, 4, 6, 7 correctly placed.
\(1\) (M1)(A1)(ft) (C2)
Notes: Award (M1)(A0) for listing the elements of their set \(B \cap A’\);shading the correct region on diagram; or an answer of \(1/7\) with a correct Venn diagram. Follow through from part (a).
Correct, from \((2,{\text{ }}2){\text{ }}(3,{\text{ }}3)\) and \((5,{\text{ }}5)\) on sample space
OR
Correct, from a labelled tree diagram
OR
Correct, from a sample space diagram
OR
Correct, from \(3 \times \frac{1}{4} \times \frac{1}{6}\;\;\;\)(or equivalent) (A1)(ft)(R1) (C2)
Notes: Do not award (A1)(ft)(R0). Award (R1) for a consistent reason with their part (a). Follow through from part (a).
Question
Peter either walks or cycles to work. The probability that he walks is 0.25. If Peter walks to work, the probability that he is late is 0.1. If he cycles to work, the probability that he is late is 0.05. The tree diagram for this information is shown.
On a day chosen at random, Peter walked to work.
Write down the probability that he was on time.[1]
For a different day, also chosen at random,
find the probability that Peter cycled to work and was late.[2]
For a different day, also chosen at random,
find the probability that, given Peter was late, he cycled to work.[3]
Answer/Explanation
Markscheme
\(0.9\) (A1) (C1)
\(0.75 \times 0.05\) (M1)
\( = 0.0375\;\;\;\left( {\frac{3}{{80}},{\text{ 3,75% }}} \right)\) (A1) (C2)
\(\frac{{0.75 \times 0.05}}{{0.75 \times 0.05 + 0.25 \times 0.1}}\) (M1)(M1)
Note: Award (M1) for their correct numerator, (M1) for their correct denominator, ie, \(\left( {\frac{{{\text{their (b)}}}}{{{\text{their (b)}} + 0.25 \times 0.1}}} \right)\).
Do not award (M1) for their \(0.0375\) or \(0.0625\) if not a correct part of a fraction.
\( = 0.6\;\;\;\left( {\frac{3}{5},{\text{ }}60\% } \right)\) (A1)(ft) (C3)
Note: Follow through from part (b).
Question
On a work day, the probability that Mr Van Winkel wakes up early is \(\frac{4}{5}\).
If he wakes up early, the probability that he is on time for work is \(p\).
If he wakes up late, the probability that he is on time for work is \(\frac{1}{4}\).
The probability that Mr Van Winkel arrives on time for work is \(\frac{3}{5}\).
Complete the tree diagram below.
[2]
Find the value of \(p\).[4]
Answer/Explanation
Markscheme
(A1)(A1) (C2)
Note: Award (A1) for each correct pair of probabilities.[2 marks]
\(\frac{4}{5}p + \frac{1}{5} \times \frac{1}{4} = \frac{3}{5}\) (A1)(ft)(M1)(M1)
Note: Award (A1)(ft) for two correct products from part (a), (M1) for adding their products, (M1) for equating the sum of any two probabilities to \(\frac{3}{5}\).
\((p = ){\text{ }}\frac{{11}}{{16}}{\text{ }}(0.688,{\text{ }}0.6875)\) (A1)(ft) (C4)
Note: Award the final (A1)(ft) only if \(0 \leqslant p \leqslant 1\). Follow through from part (a).[4 marks]
Question
In the Canadian city of Ottawa:
\[\begin{array}{*{20}{l}} {{\text{97% of the population speak English,}}} \\ {{\text{38% of the population speak French,}}} \\ {{\text{36% of the population speak both English and French.}}} \end{array}\]
The total population of Ottawa is \(985\,000\).
Calculate the percentage of the population of Ottawa that speak English but not French.[2]
Calculate the number of people in Ottawa that speak both English and French.[2]
Write down your answer to part (b) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10\) and k \( \in \mathbb{Z}\).[2]
Answer/Explanation
Markscheme
\(97 – 36\) (M1)
Note: Award (M1) for subtracting 36 from 97.
OR
(M1)
Note: Award (M1) for 61 and 36 seen in the correct places in the Venn diagram.
\( = 61{\text{ }}(\% )\) (A1) (C2)
Note: Accept 61.0 (%).[2 marks]
\(\frac{{36}}{{100}} \times 985\,000\) (M1)
Note: Award (M1) for multiplying 0.36 (or equivalent) by \(985\,000\).
\( = 355\,000{\text{ }}(354\,600)\) (A1) (C2)[2 marks]
\(3.55 \times {10^5}{\text{ }}(3.546 \times {10^5})\) (A1)(ft)(A1)(ft) (C2)
Note: Award (A1)(ft) for 3.55 (3.546) must match part (b), and (A1)(ft) \( \times {10^5}\).
Award (A0)(A0) for answers of the type: \(35.5 \times {10^4}\). Follow through from part (b).[2 marks]
Question
The Home Shine factory produces light bulbs, 7% of which are found to be defective.
Francesco buys two light bulbs produced by Home Shine.
The Bright Light factory also produces light bulbs. The probability that a light bulb produced by Bright Light is not defective is \(a\).
Deborah buys three light bulbs produced by Bright Light.
Write down the probability that a light bulb produced by Home Shine is not defective.[1]
Find the probability that both light bulbs are not defective.[2]
Find the probability that at least one of Francesco’s light bulbs is defective.[2]
Write down an expression, in terms of \(a\), for the probability that at least one of Deborah’s three light bulbs is defective.[1]
Answer/Explanation
Markscheme
0.93 (93%) (A1) (C1)[1 mark]
\(0.93 \times 0.93\) (M1)
Note: Award (M1) for squaring their answer to part (a).
0.865 (0.8649; 86.5%) (A1)(ft) (C2)
Notes: Follow through from part (a).
Accept \(0.86{\text{ }}\left( {{\text{unless it follows }}\frac{{93}}{{100}} \times \frac{{92}}{{99}}} \right)\).[2 marks]
\(1 – 0.8649\) (M1)
Note: Follow through from their answer to part (b)(i).
OR
\(0.07 \times 0.07 + 2 \times (0.07 \times 0.93)\) (M1)
Note: Follow through from part (a).
0.135 (0.1351; 13.5%) (A1)(ft) (C2)[2 marks]
\(1 – {a^3}\) (A1) (C1)
Note: Accept \(3{a^2}(1 – a) + 3a{(1 – a)^2} + {(1 – a)^3}\) or equivalent.[1 mark]
Question
Rosewood College has 120 students. The students can join the sports club (\(S\)) and the music club (\(M\)).
For a student chosen at random from these 120, the probability that they joined both clubs is \(\frac{1}{4}\) and the probability that they joined the music club is\(\frac{1}{3}\).
There are 20 students that did not join either club.
Complete the Venn diagram for these students.
[2]
One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.[2]
Determine whether the events \(S\) and \(M\) are independent.[2]
Answer/Explanation
Markscheme
(A1)(A1) (C2)
Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.[2 marks]
\(\frac{{30}}{{90}}{\text{ }}\left( {\frac{1}{3},{\text{ }}0.333333 \ldots ,{\text{ }}33.3333 \ldots \% } \right)\) (A1)(ft)(A1)(ft) (C2)
Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.[2 marks]
\({\text{P}}(S) \times {\text{P}}(M) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}\) (R1)
Note: Award (R1) for multiplying their by \(\frac{1}{3}\).
therefore the events are independent \(\left( {{\text{as P}}(S \cap M) = \frac{1}{4}} \right)\) (A1)(ft) (C2)
Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.
Do not award (R0)(A1)(ft).
Do not award final (A1) if \({\text{P}}(S) \times {\text{P}}(M)\) is not calculated. Follow through from part (a).[2 marks]
Question
In an international competition, participants can answer questions in only one of the three following languages: Portuguese, Mandarin or Hindi. 80 participants took part in the competition. The number of participants answering in Portuguese, Mandarin or Hindi is shown in the table.
A boy is chosen at random.
State the number of boys who answered questions in Portuguese.[1]
Find the probability that the boy answered questions in Hindi.[2]
Two girls are selected at random.
Calculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.[3]
Answer/Explanation
Markscheme
20 (A1) (C1)[1 mark]
\(\frac{5}{{43}}\,\,\,\left( {0.11627 \ldots ,\,\,11.6279 \ldots {\text{% }}} \right)\) (A1)(A1) (C2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.[2 marks]
\(\frac{7}{{37}} \times \frac{{12}}{{36}} + \frac{{12}}{{37}} \times \frac{7}{{36}}\) (A1)(M1)
Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.
\( = \frac{{14}}{{111}}\,\,\left( {\,0.12612 \ldots ,\,\,12.6126\,{\text{% }}} \right)\) (A1) (C3)[3 marks]
Question
Dune Canyon High School organizes its school year into three trimesters: fall/autumn (\(F\)), winter (\(W\)) and spring (\(S\)). The school offers a variety of sporting activities during and outside the school year.
The activities offered by the school are summarized in the following Venn diagram.
Write down the number of sporting activities offered by the school during its school year.[1]
Determine whether rock-climbing is offered by the school in the fall/autumn trimester.[1]
Write down the elements of the set \(F \cap W’\);[1]
Write down \(n(W \cap S)\).[1]
Write down, in terms of \(F\), \(W\) and \(S\), an expression for the set which contains only archery, baseball, kayaking and surfing.[2]
Answer/Explanation
Markscheme
15 (A1) (C1)[1 mark]
no (A1) (C1)
Note: Accept “it is only offered in Winter and Spring”.[1 mark]
volleyball, golf, cycling (A1) (C1)
Note: Responses must list all three sports for the (A1) to be awarded.[1 mark]
4 (A1) (C1)[1 mark]
\((F \cup W \cup S)’\)\(\,\,\,\)OR\(\,\,\,\)\(F’ \cap W’ \cap S’\) (or equivalent) (A2) (C2)[2 marks]
Question
Sara regularly flies from Geneva to London. She takes either a direct flight or a non-directflight that goes via Amsterdam.
If she takes a direct flight, the probability that her baggage does not arrive in London is 0.01.
If she takes a non-direct flight the probability that her baggage arrives in London is 0.95.
The probability that she takes a non-direct flight is 0.2.
Complete the tree diagram.[3]
Find the probability that Sara’s baggage arrives in London.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair of probabilities.[3 marks]
\(0.8 \times 0.99 + 0.2 \times 0.95\) (A1)(ft)(M1)
Note: Award (A1)(ft) for two correct products of probabilities taken from their diagram, (M1) for the addition of their products.
\( = 0.982{\text{ }}\left( {98.2\% ,{\text{ }}\frac{{491}}{{500}}} \right)\) (A1)(ft) (C3)
Note: Follow through from part (a).[3 marks]
Question
All the children in a summer camp play at least one sport, from a choice of football (\(F\)) or basketball (\(B\)). 15 children play both sports.
The number of children who play only football is double the number of children who play only basketball.
Let \(x\) be the number of children who play only football.
There are 120 children in the summer camp.
Write down an expression, in terms of \(x\), for the number of children who play only basketball.[1]
Complete the Venn diagram using the above information.
[2]
Find the number of children who play only football.[2]
Write down the value of \(n(F)\).[1]
Answer/Explanation
Markscheme
\(\frac{1}{2}x\) (A1) (C1)[1 mark]
(A1)(A1)(ft) (C2)
Notes: Award (A1) for 15 placed in the correct position, award (A1)(ft) for \(x\) and their \(\frac{1}{2}x\) placed in the correct positions of diagram. Do not penalize the absence of 0 inside the rectangle and award at most (A1)(A0) if any value other than 0 is seen outside the circles. Award at most (A1)(A0) if 35 and 70 are seen instead of \(x\) and their \(\frac{1}{2}x\).[2 marks]
\(x + \frac{1}{2}x + 15 = 120\) or equivalent (M1)
Note: Award (M1) for adding the values in their Venn and equating to 120 (or equivalent).
\((x = ){\text{ }}70\) (A1)(ft) (C2)
Note: Follow through from their Venn diagram, but only if the answer is a positive integer and \(x\) is seen in their Venn diagram.[2 marks]
85 (A1)(ft) (C1)
Note: Follow through from their Venn diagram and their answer to part (c), but only if the answer is a positive integer and less than 120.[1 mark]
Question
The Home Shine factory produces light bulbs, 7% of which are found to be defective.
Francesco buys two light bulbs produced by Home Shine.
The Bright Light factory also produces light bulbs. The probability that a light bulb produced by Bright Light is not defective is \(a\).
Deborah buys three light bulbs produced by Bright Light.
Write down the probability that a light bulb produced by Home Shine is not defective.[1]
Find the probability that both light bulbs are not defective.[2]
Find the probability that at least one of Francesco’s light bulbs is defective.[2]
Write down an expression, in terms of \(a\), for the probability that at least one of Deborah’s three light bulbs is defective.[1]
Answer/Explanation
Markscheme
0.93 (93%) (A1) (C1)[1 mark]
\(0.93 \times 0.93\) (M1)
Note: Award (M1) for squaring their answer to part (a).
0.865 (0.8649; 86.5%) (A1)(ft) (C2)
Notes: Follow through from part (a).
Accept \(0.86{\text{ }}\left( {{\text{unless it follows }}\frac{{93}}{{100}} \times \frac{{92}}{{99}}} \right)\).[2 marks]
\(1 – 0.8649\) (M1)
Note: Follow through from their answer to part (b)(i).
OR
\(0.07 \times 0.07 + 2 \times (0.07 \times 0.93)\) (M1)
Note: Follow through from part (a).
0.135 (0.1351; 13.5%) (A1)(ft) (C2)[2 marks]
\(1 – {a^3}\) (A1) (C1)
Note: Accept \(3{a^2}(1 – a) + 3a{(1 – a)^2} + {(1 – a)^3}\) or equivalent.[1 mark]
Question
Rosewood College has 120 students. The students can join the sports club (\(S\)) and the music club (\(M\)).
For a student chosen at random from these 120, the probability that they joined both clubs is \(\frac{1}{4}\) and the probability that they joined the music club is\(\frac{1}{3}\).
There are 20 students that did not join either club.
Complete the Venn diagram for these students.
[2]
One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.[2]
Determine whether the events \(S\) and \(M\) are independent.[2]
Answer/Explanation
Markscheme
(A1)(A1) (C2)
Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.[2 marks]
\(\frac{{30}}{{90}}{\text{ }}\left( {\frac{1}{3},{\text{ }}0.333333 \ldots ,{\text{ }}33.3333 \ldots \% } \right)\) (A1)(ft)(A1)(ft) (C2)
Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.[2 marks]
\({\text{P}}(S) \times {\text{P}}(M) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}\) (R1)
Note: Award (R1) for multiplying their by \(\frac{1}{3}\).
therefore the events are independent \(\left( {{\text{as P}}(S \cap M) = \frac{1}{4}} \right)\) (A1)(ft) (C2)
Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.
Do not award (R0)(A1)(ft).
Do not award final (A1) if \({\text{P}}(S) \times {\text{P}}(M)\) is not calculated. Follow through from part (a).[2 marks]
Question
A group of 60 sports enthusiasts visited the PyeongChang 2018 Winter Olympic games to watch a variety of sporting events.
The most popular sports were snowboarding (S), figure skating (F) and ice hockey (H).
For this group of 60 people:
4 did not watch any of the most popular sports,
x watched all three of the most popular sports,
9 watched snowboarding only,
11 watched figure skating only,
15 watched ice hockey only,
7 watched snowboarding and figure skating,
13 watched figure skating and ice hockey,
11 watched snowboarding and ice hockey.
Complete the Venn diagram using the given information.
[3]
Find the value of x.[2]
Write down the value of \(n\left( {\left( {F \cup H} \right) \cap S’} \right)\).[1]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for 4 in correct place.
Award (A1) for 9, 11, 15 in correct place.
Award (A1) for 7 − x, 13 − x, 11 − x in correct place.
Accept 2, 8 and 6 in place of 7 − x, 13 − x, 11 − x.[3 marks]
\(4 + 9 + 11 + 15 + x + \left( {7 – x} \right) + \left( {11 – x} \right) + \left( {13 – x} \right) = 60\) (M1)
Note: Award (M1) for equating the sum of at least seven of the entries in their Venn diagram to 60.
\(\left( {x = } \right)\,\,5\) (A1)(ft) (C2)
Note: Follow through from part (a), but only if answer is positive.[2 marks]
34 (A1)(ft) (C1)
Note: Follow through from their Venn diagram.[1 mark]
Question
A group of 30 students were asked about their favourite topping for toast.
18 liked peanut butter (A)
10 liked jam (B)
6 liked neither
Show this information on the Venn diagram below.
[2]
Find the number of students who like both peanut butter and jam.[2]
Find the probability that a randomly chosen student from the group likes peanut butter, given that they like jam.[2]
Answer/Explanation
Markscheme
OR 
(A2) (C2)
Note: Award (A2) for 3 correctly placed values, and no extras (4 need not be seen), (A1) for 2 correctly placed values, (A0) for 1 or no correctly placed values.[2 marks]
18 + 10 + 6 = 30 (M1)
= 4 (A1) (C2)[2 marks]
\({\text{P}}(A|B) = \frac{4}{{10}}\left( {\frac{2}{5},{\text{ }}0.4,{\text{ }}40{\text{ }}\% } \right)\) (A1)(ft)(A1) (C2)
Note: Award (A1)(ft) for their numerator from part (b), (A1) for denominator.[2 marks]
[MAI 4.9] DISCRETE DISTRIBUTIONS-neha
Question
The probability distribution of the discrete random variable \(X\) is given by the table
Find
(a) the expected value \(E(X)\) of \(X\).
(b) the mode of \(X\).
(c) the median of \(X\).
(d) the lower quartile \(Q_{1}\) and the upper quartile \(Q_{3}\).
Answer/Explanation
Ans
(a) \(2.35\) (b) mode = \(1\) (c) median = \(2\) (d) \(Q_{1} = 1 \ Q_{3} = 3.5\)
Question
The probability distribution of the discrete random variable \(X\) is given by the table
Find the values of \(a\) and \(b\) given that \(E(X)\) = \(2.2\)
Answer/Explanation
Ans
\(a = 0.2\) \(b = 0.4\)
Question
The probability distribution of the discrete random variable \(X\) is given by the table
Nikos selects a number at random.
If he selects \(1\) he earns \(10\) €. If he selects \(2\) he earns \(5\) €. If he selects \(3\) he loses \(4\) €
(a) Find the expected value of \(X\).
(b) Find the expected value of the profit for Nikos. Is the game fair?
Answer/Explanation
Ans
\(E(X) = 2.2\) E(Profit) = \(0\), so it is fair.
Question
A discrete random variable \(X\) has its probability distribution given by
\(P(X = x)=\frac{x}{6}\), where x is \(1, 2, 3\).
(a) Complete the following table showing the probability distribution for \(X\) .
(b) Find \(E(X)\).
Answer/Explanation
Ans
(a) 
(b) Find E(\(X\)) = \(7/3\)
Question
Each of the following \(10\) words is placed on a card and put in a hat.
ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN
We pick a card at random. Let \(X\) be the size (number of letters) of the corresponding word.
(a) Give the probability distribution for \(X\) (i.e. the table of probabilities)
(b) Find the expected number of \(X\).
Answer/Explanation
Ans
(a) 
(b) Find E(\(X\)) = \(3.9\)
Question
The probability distribution of the discrete random variable \(X\) is given by the following table.
(a) Find the value of \(p\).
(b) Calculate the expected value of \(X\).
Answer/Explanation
Ans
(a) \((0.4 + p + 0.2 + 0.07 + 0.02 = 1)\), \(\Rightarrow p = 0.31\)
(b) \(E(X) = 1(0.4) + 2(0.31) + 3(0.2) + 4(0.07) + 5(0.02) = 2\)
Question
A discrete random variable \(X\) has a probability distribution as shown in the table below.
(a) Find the value of \(a + b\).
(b) Given that \(E(X) =1.5\), find the value of \(a\) and of \(b\).
Answer/Explanation
Ans
(a) \(0.1 + a + 0.3 + b = 1\) \(\Rightarrow a + b = 0.6\)
(b) \(0 \times 0.1 + 1 \times a + 2 \times 0.3 + 3 \times b\)
\(0 + a + 0.6 + 3b = 1.5\)
\(a + 3b = 0.9\)
Solving simultaneously gives
\(a = 0.45\) \(b = 0.15\)
Question
The table below shows the probability distribution of a discrete random variable \(X\) .
Given that \(E(X) =1.55\), find the value of \(a\) and of \(b\).
Answer/Explanation
Ans
\( 0.2 + a + b + 0.25 = 1(a + b = 0.55)\)
\(E(X) = a+2b+0.75 = 1.55\)
\(\Rightarrow a + 2b = 0.8\)
\(a = 0.3\) and \(b = 0.25\)
Question
In a game a player rolls a biased tetrahedral (four-faced) die. The probability of each possible score is shown below.
Find the probability of a total score of six after two rolls.
Answer/Explanation
Ans
\(\sum_{all\ x}P(X = x)=1 \Rightarrow \frac{1}{5}+\frac{2}{5}+\frac{1}{10} + x =1 \Rightarrow x = \frac{3}{10}\)
P(scoring six after two rolls) = \(\left ( \frac{1}{10}\times \frac{1}{10} \right )+ 2 \times \left ( \frac{2}{5}\times \frac{3}{10} \right )= \frac{1}{4}\)
Question
The following table shows the probability distribution of a discrete random variable \(X\).
(a) Find the value of \(k\).
(b) Find the expected value of \(X\).
Answer/Explanation
Ans
(a) \(10k^{2}+3k+0.6=1 \Rightarrow 10k^{2}+3k-0.4=0 \Leftrightarrow k=0.1\) (by GDC)
(b) E(X) = \(- 1 \times 0.2 + 2 \times 0.4 + 3 \times 0.3 = 1.5\)
Question
A discrete random variable \(X\) has its probability distribution given by
\(P(X = x) = k(x + 1)\), where \(x\) is \(0, 1, 2, 3, 4\).
(a) Complete the following table showing the probability distribution for \(X\) (in terms of \(k\))
(b) Show that \(k = \frac{1}{15}\).
(c) Find \(E(X)\).
Answer/Explanation
Ans
(a) 
(b) \(k \times 1 + k \times 2 + k \times 3 + k \times 4 + k \times 5 = 15k = 1 \Leftrightarrow k = 15\)
(c) E(X) = \(0 \times \frac{1}{15}+ 1 \times \frac{2}{15}+ 2 \times \frac{3}{15}+ 3 \times \frac{4}{15}+ 4 \times \frac{5}{15} = \frac{40}{15} = \frac{8}{3}\)
Question
The probability distribution of a discrete random variable \(X\) is defined by
\(P(X = x) = cx(5 − x)\), \(x = 1, 2, 3, 4\).
(a) Find the value of \(c\).
(b) Find \(E(X)\).
Answer/Explanation
Ans
(a) \(\sum P(X = x) = 1 \Rightarrow 4c + 6c + 6c + 4c = 1 \Rightarrow 20c = 1 \Rightarrow c = \frac{1}{20} (=0.05)\)
(b) \(E(X) = \sum xP(X = x)= (1\times 0.2)+(2\times 0.3)+(3\times 0.3)+(4\times 0.2) = 2.5\)
Question
The probability distribution of a discrete random variable \(X\) is given by
\(P(X=x)=k\left ( \frac{2}{3} \right )^{x}\), for \(x = 0,1, 2, ……\)
Find the value of \(k\).
Answer/Explanation
Ans
\(\sum_{all \ x}P(X =x)=1\Rightarrow k+\frac{2}{3}k+\left ( \frac{2}{3} \right )^{2}k+\left ( \frac{2}{3} \right )^{3}k+…………..= 1\)
\(k\left ( \frac{1}{1-\frac{2}{3}} \right )=1\Rightarrow k = \frac{1}{3}\)
Question
Three students, Kim, Ching Li and Jonathan each have a pack of cards, from which they select a card at random. Each card has a \(0, 3, 4\), or \(9\) printed on it.
(a) Kim states that the probability distribution for her pack of cards is as follows.
Explain why Kim is incorrect.
(b) Ching Li correctly states that the probability distribution for her pack of cards is as follows.
Find the value of \(k\).
(c) Jonathan correctly states that the probability distribution for his pack of cards is given by \(P(X=x)=\frac{x+1}{20}\) .One card is drawn at random from his pack. Calculate the
probability that the number on the card drawn
(i) is \(0\). (ii) is greater than \(0\).
Answer/Explanation
Ans
(a) Sum = \(1.3\) which is greater than \(1\)
(b) \(3k + 0.7 = 1 \Rightarrow k = 0.1\)
(c) (i) \(P(X=0) = \frac{0+1}{20}=\frac{1}{20}
(ii) \(P(X>0) = 1-P(X=0)=\frac{19}{20} \left ( or \frac{4}{20}+\frac{5}{20}+\frac{10}{20}\right ) = \frac{19}{20}\)
Notice: in fact, the probability distribution is
Question
A biased die with four faces is used in a game. A player pays 10 counters to roll the die. The table below shows the
possible scores on the die, the probability of each score and the number of counters the player receives in return for each
score.
(a) The player throws the die twice. Find the probability that
(i) he has a total score of \(3\). (ii) he has a total score of \(4\).
(b) Find the value of \(n\) in order for the player to get an expected return of \(9\) counters per roll.
Answer/Explanation
Ans
(a) (i) \(1+2\) or \(2+1\) Prob = \(\frac{1}{2}\times \frac{1}{5}\times 2 = \frac{1}{5}\)
(ii) \(1+3\) or \(3+1\) or \(2+2\) \(Prob = \frac{1}{2}\times \frac{1}{5}\times 2 + \frac{1}{5}\times \frac{1}{5}= \frac{1}{5} + \frac{1}{25} = \frac{6}{25}\)
(b) Let \(X\) be the number of counters the player receives in return.
\(E(X) = \sum p(x) \times x = 9\)
\(\Leftrightarrow \left ( \frac{1}{2}\times 4 \right ) + \left ( \frac{1}{5}\times 5 \right ) + \left ( \frac{1}{5}\times 15 \right ) + \left ( \frac{1}{10}\times n \right ) = 9\)
\(\Leftrightarrow \frac{1}{10}n = 3 \Leftrightarrow n = 30\)
Question
Two fair four-sided dice, one red and one green, are thrown. For each die, the faces are labelled \(1, 2, 3, 4\). The score for each die is the number which lands face down.
The sample space is shown below:
(a) Write down the probability that two scores of \(4\) are obtained.
Let \(X\) be the number of \(4\)s that land face down.
(b) Complete the following probability distribution for \(X\).
(c) Find \(E(X)\).
Chris plays a game where he rolls the dice above.
If two \(4\)s are obtained he wins \(20\)€.
If only one \(4\) is obtained he wins \(5\)€.
If no \(4\) is obtained he loses \(2\)€
(d) Find the expected amount earned in one game.
(e) If Chris plays this game 100 times find the amount he is expected to win.
(f) If Chris plays this game twice find the probability that he earns \(18\)€.
Answer/Explanation
Ans
(a) Probability of two \(4\)s is \(\frac{1}{16} (=0.0625)\)
(b) 
(c) \(E(X) = \sum_{0}^{2}xP(X = x), = 0\times \frac{9}{16} + 1 \times \frac{6}{16} +2 \times \frac{1}{16}\frac{8}{16}\left ( =\frac{1}{2} \right )\)
or \(E(X) = np = 2 \times \frac{1}{4} = \frac{1}{2}\)
(d) 
Expected amount = \(-2\times \frac{9}{16} + 5\times \frac{6}{16} + 20\times \frac{1}{16} = 2\)€
(e) \(100 \times 2 = 200\)€
(f) \(18\) € implies \(0\) fours and \(2\) fours or vice versa
\(\frac{9}{16} \times \frac{1}{16} \times 2 = \frac{9}{128}\)
Question
Bag A contains \(2\) red balls and \(3\) green balls. Two balls are chosen at random from the bag without replacement. Let \(X\) denote the number of red balls chosen. The following table
shows the probability distribution for \(X\)
(a) Calculate \(E(X)\), the mean number of red balls chosen.
Bag B contains \(4\) red balls and \(2\) green balls. Two balls are chosen at random from bag B.
(b) (i) Draw a tree diagram to represent the above information, including the probability of each event.
(ii) Hence find the probability distribution for \(Y\), where \(Y\) is the number of red balls chosen.
A standard die with six faces is rolled. If a \(1\) or \(6\) is obtained, two balls are chosen from bag A, otherwise two balls are chosen from bag B.
(c) Calculate the probability that two red balls are chosen.
(d) Given that two red balls are obtained, find the conditional probability that a \(1\) or \(6\) was rolled on the die.
Answer/Explanation
Ans
(a) \(E(X) = 0\times \frac{3}{10} + 1\times \frac{6}{10} + 2 \times \frac{1}{10} = \frac{8}{10} (0.8)\)
(b) (i) 
(ii) \(P(Y = 0) = \frac{2}{5}\times \frac{1}{5} = \frac{2}{30}\)
\(P(Y =1) = P(RG) + P(GR) \left ( = \frac{4}{6}\times \frac{2}{5} + \frac{2}{6}\times \frac{4}{5}\right ) = \frac{16}{30}\)
\(P(Y =2) = \frac{4}{6}\times \frac{3}{5} = \frac{12}{30}\)
Forming a distribution
(c) \(P(RR) = \frac{1}{3}\times \frac{1}{10} + \frac{2}{3}\times \frac{12}{30} = \frac{27}{90} \left ( \frac{3}{10}, 0.3 \right )\)
(d) P(\(1\) or \(6\)|RR) = P(A|RR) = \(\frac{P(A\cap RR)}{P(RR)} = \frac{1}{30}\div \frac{27}{90} = \frac{3}{27} \left ( \frac{1}{9}, 0.111 \right )\)
Question
A four-sided die has three blue faces and one red face. The die is rolled.
Let \(B\) be the event a blue face lands down, and \(R\) be the event a red face lands down.
(a) Write down the values of
(i) \(P(B)\)
(ii) \(P(R)\)
(b) If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is represented by the following tree diagram, where \(p\), \(s\), \(t\) are probabilities.
Find the value of \(p\), of \(s\) and of \(t\).
Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores \(2\) and is finished. If the red face lands down, he scores \(1\) and rolls one more time. Let \(X\) be the total score obtained.
(c) (i) Show that \(P (X = 3)\) = \(\frac{3}{16}\).
(ii) Find \(P (X = 2)\).
(d) (i) Construct a probability distribution table for \(X\).
(ii) Calculate the expected value of \(X\).
(e) If the total score is \(3\), Guiseppi wins \($ 10\). If the total score is \(2\), Guiseppi gets nothing. He plays the game twice. Find the probability that he wins exactly \($ 10\).
Answer/Explanation
Ans
(a) (i) P(B) = \(\frac{3}{4}\) (ii) P(R) = \(\frac{1}{4}\)
(b) \(p = \frac{3}{4}\), \(s = \frac{1}{4}\), \(t = \frac{3}{4}\)
(c) (i) \(P(X = 3)\) = P (getting 1 and 2) = \(\frac{1}{4}\times \frac{3}{4} = \frac{3}{16}\)
(ii) \(P(X = 2) = \frac{1}{4}\times \frac{1}{4} + \frac{3}{4} \left ( or 1 -\frac{3}{16} \right ) = \frac{13}{16}\)
(d) (i) 
(ii) \(E(X) = 2\frac{13}{16} + 3\frac{3}{16} = \frac{35}{16}\)
(e) win $\(10\) \(\Rightarrow\) scores \(3\) one time, \(2\) other time
P(win $\(10\)) = \(P(3)\times P(2)+ P(2)\times P(3) = 2\left ( \frac{13}{16}\times \frac{3}{16} \right ) = \frac{78}{256} \left ( = \frac{39}{128} \right )\)
Question
John removes the labels from three cans of tomato soup and two cans of chicken soup in order to enter a competition, and puts the cans away. He then discovers that the cans are identical, so
that he cannot distinguish between cans of tomato soup and chicken soup. Some weeks later he decides to have a can of chicken soup for lunch. He opens the cans at random until he opens a
can of chicken soup. Let \(Y\) denote the number of cans he opens. Find
(a) the possible values of \(Y\),
(b) the probability of each of these values of \(Y\),
(c) the expected value of \(Y\).
(d) Write down the probability distribution of \(X\) if \(X\) denotes the number of cans he opens until he opens a can of tomato soup.
Answer/Explanation
Ans
(a) \(1, 2, 3, 4\)
(b) P(Y = 1) = \(\frac{2}{5}\)
P(Y = 2) = \(\frac{3}{5} \times \frac{2}{4}\) = \(\frac{3}{10}\)
P(Y = 3) = \(\frac{3}{5} \times \frac{2}{4} \times \frac{2}{3}\) = \(\frac{1}{5}\)
P(Y = 4) = \(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} \) = \(\frac{1}{10}\)
(c) \(E(Y) = 1 \times \frac{2}{5} + 2 \times \frac{3}{10} + 3\times \frac{1}{5} + 4\times \frac{1}{10} = 2\)
(d) 
[MAI 4.10] BINOMIAL DISTRIBUTION-loyola
Question
[Maximum mark: 10]
A fair coin is tossed eight times.
(a) Calculate the probability of obtaining
| exactly 4 heads | |
| exactly 3 heads | |
| 3, 4 or 5 heads | |
| no heads | |
| always heads | |
| at most 2 heads | |
| at least 3 heads |
(b) Find the expected number of heads and the variance of the number of heads
| E(X) | Var(X) |
Answer/Explanation
Answer:
(a)
| exactly 4 heads | 0.273 |
| exactly 3 heads | 0.219 |
| 3, 4 or 5 heads | 0.711 |
| no heads | 0.00391 |
| always heads | 0.00391 |
| at most 2 heads | 0.145 |
| at least 3 heads | 0.855 |
(b)
| E(X) | 4 | Var(X) | 2 |
Question
[Maximum mark: 6]
The random variable X follows the binomial distribution B(n,p). Given that E(X) = 10 and
Var(X) = 6 find the values of n and p.
Answer/Explanation
Answer:
np =10 and np(1- p) = 6 . Hence 10(1- p) = 6 ⇔ p = 0.4 and n = 25
Question
[Maximum mark: 4]
A fair coin is tossed five times. Calculate the probability of obtaining
(a) exactly three heads; [2]
(b) at least one head. [2]
Answer/Explanation
Answer:
B(n, p) with n = 5 and \(p=\frac{1}{2}\)
(a) P(X = 3) = 0.3125 … = 0.313
(b) P(X ≥ 1) = 0.969
Question
[Maximum mark: 4]
The probability of obtaining heads on a biased coin is 0.18. The coin is tossed seven times.
(a) Find the probability of obtaining exactly two heads. [2]
(b) Find the probability of obtaining at least two heads. [2]
Answer/Explanation
Answer:
B(n, p) with n = 7 and p = 0.18
(a) P(X = 2) = 0.252
(b) P(X ≥ 2) = 0.368
Question
[Maximum mark: 5]
A factory makes switches. The probability that a switch is defective is 0.04.
The factory tests a random sample of 100 switches.
(a) Find the mean number of defective switches in the sample. [1]
(b) Find the probability that there are exactly six defective switches in the sample. [2]
(c) Find the probability that there is at least one defective switch in the sample. [2]
Answer/Explanation
Answer:
B(n, p) with n = 100 and p = 0.04
(a) mean = np = 100 x 0.04 = 4
(b) P(X = 6) = 0.105
(c) P(X ≥ 1) = 0.983
Question
[Maximum mark: 5]
A factory makes calculators. Over a long period, 2 % of them are found to be faulty. A random sample of 100 calculators is tested.
(a) Write down the expected number of faulty calculators in the sample. [1]
(b) Find the probability that three calculators are faulty. [2]
(c) Find the probability that more than one calculator is faulty. [2]
Answer/Explanation
Answer:
X ~ B(100,0.02)
(a) E(X) = 100 × 0.02 = 2
(b) (i) P(X = 3) = 0.182 (ii) P(X > 1) = 0.597
Question
[Maximum mark: 7]
A box contains 35 red discs and 5 black discs. A disc is selected at random and its colour noted. The disc is then replaced in the box.
(a) In eight such selections, what is the probability that a black disc is selected
(i) exactly once?
(ii) at least once? [5]
(b) The process of selecting and replacing is carried out 400 times.
What is the expected number of black discs that would be drawn? [2]
Answer/Explanation
Answer:
\(p(Red)=\frac{35}{40}=\frac{7}{8}\\) \(p(Black)=\frac{5}{40}=\frac{1}{8}\)
(a) B(n, p) with n = 5, \(p=\frac{1}{8}\)
(i) p(one black) = P(X = 1) = 0.393 to 3 s.f. (ii) p(at least one black) = P(X ≥ 1) = 0.656
(b) 400 draws: expected number of blacks \(=\frac{400}{8}=50\)
Question
[Maximum mark: 4]
In a school, \(\frac{1}{3}\) of the students travel to school by bus. Five students are chosen at random.
(a) Find the probability that exactly 3 of them travel to school by bus. [2]
(b) Find the probability that at most 3 of them travel to school by bus. [2]
Answer/Explanation
Answer:
X ~ B(n, p) with n = 5 and \(p=\frac{1}{3}\)
Therefore P(X = 3) = 0.165
Question
[Maximum mark: 4]
When John throws a stone at a target, the probability that he hits the target is 0.4. He throws a stone 6 times.
(a) Find the probability that he hits the target exactly 4 times. [2]
(b) Find the probability that he hits the target for the first time on his third throw. [2]
Answer/Explanation
Answer:
(a) Probability = 0.138
(b) Probability = (0.6)2 × 0.4 = 0.144 \(\left ( or\frac{18}{125} \right )\)
Question
[Maximum mark: 4]
When a boy plays a game at a fair, the probability that he wins a prize is 0.25. He plays the game 10 times. Let X denote the total number of prizes that he wins. Assuming that the games are independent, find
(a) E(X) [2]
(b) P (X ≤ 2). [2]
Answer/Explanation
Answer:
(a) X is B(10, 0.25)
E(X) = 10 × 0.25 = 2.5
(b) P(X ≤ 2) = 0.526
Question
[Maximum mark: 3]
On a television channel the news is shown at the same time each day. The probability that Alice watches the news on a given day is 0.4. Calculate the probability that on five consecutive days, she watches the news on at most three days.
Answer/Explanation
Answer:
X is Binomial n = 5 p = 0.4
P(X ≤ 3) = 0.913 to 3 s.f.
Question
[Maximum mark: 9]
The probability of obtaining heads on a biased coin is \(\frac{1}{3}\).
(a) Sammy tosses the coin three times. Find the probability of getting
(i) three heads;
(ii) two heads and one tail. [4]
(b) Amir plays a game in which he tosses the coin 12 times.
(i) Find the expected number of heads.
(ii) Amir wins $10 for each head obtained and loses $ 6 for each tail. Find his expected winnings. [5]
Answer/Explanation
Answer:
(a) B(n, p) with n = 3, \(p=\frac{1}{3}\)
(i) P(X = 3) = 0.0370 or P(3H) \(=\left ( \frac{1}{3} \right )^{3}=\frac{1}{27}\)
(ii) P(X = 3) = 0.222 or P(2H, 1T) \(=3\frac{1}{3}^{2}\frac{2}{3}=\frac{2}{9}\)
(b) (i) expected number of heads = np \(=\left ( \frac{1}{3}\times 12 \right )=4\)
(ii) 4 heads, so 8 tails
E(winnings) = 4 × 10 – 8 × 6 (= 40 – 48) = –$ 8
Question
[Maximum mark: 5]
A biology test consists of seven multiple choice questions. Each question has five possible answers, only one of which is correct. At least four correct answers are required to pass the test. Juan does not know the answer to any of the questions so, for each question, he selects the answer at random.
(a) Find the probability that Juan answers exactly four questions correctly. [3]
(b) Find the probability that Juan passes the biology test. [2]
Answer/Explanation
Answer:
B(n, p) with n = 7, \(p=\frac{1}{5}\)
P(X = 4) = 0.0287
P(X ≥ 4) = 0.0333
Question
[Maximum mark: 6]
In an examination of 20 multiple-choice questions each question has four possible answers, only one of which is correct. Robert randomly guesses the answer to each question.
(a) Find his expected number of correct answers. [2]
(b) Find the probability that Robert obtains this expected number of correct answers. [2]
(c) Find the probability that Robert obtains less than five correct answers. [2]
Answer/Explanation
Answer:
B(n, p) with n = 20, \(p=\frac{1}{4}\)
(a) \(E(X)=20\times \frac{1}{4}=5\)
(b) P(X = 5) = 0.202 to 3 s.f.
(c) P(X < 5) = 0.415 to 3 s.f. [less than five means P(X ≤ 4) ]
Question
[Maximum mark: 4]
A satellite relies on solar cells for its power and will operate provided that at least one of the cells is working. Cells fail independently of each other, and the probability that an individual cell fails within one year is 0.8.
(a) For a satellite with ten solar cells, find the probability that all ten cells fail within one year. [2]
(b) For a satellite with ten solar cells, find the probability that the satellite is still operating at the end of one year. [2]
Answer/Explanation
Answer:
(a) P(all ten cells fail) = 0.107 (or 0.810)
(b) (satellite is still operating at the end of one year if X ≥ 1
P(X ≥ 1) = 0.893 (or 1 – 0.107= 0.893 )
Question
[Maximum mark: 4]
Let X ~ B(10,0.4). Find for X
(i) the mean (ii) the mode (iii) the variance (iv) the standard deviation.
Answer/Explanation
Answer:
(i) mean = 10 × 0.4 = 4
(ii) check P(X = 3) = 0.214, P(X = 4) =0.251, P(X = 5) = 0.201 so mode = 4
(iii) variance = 10 × 0.4 × 0.6 = 2.4
(iv) \(st. dev=\sqrt{2.4}=1.55\)
Question
[Maximum mark: 4]
\(Let X\sim B(10,\frac{1}{4})\). Find for X
(i) the mean (ii) the mode (iii) the variance (iv) the standard deviation.
Answer/Explanation
Answer:
(i) \(mean=10\times \frac{1}{4}=2.5\)
(ii) check P(X = 2) = 0.281, P(X = 3) =0.250 so mode = 2
(iii) \(variance=10\times \frac{1}{4}\times \frac{3}{4}=\frac{15}{8}=1.875\)
(iv) \(st. dev=\sqrt{1.875}=1.37\)
Question
[Maximum mark: 12]
A pair of fair dice is thrown.
(a) Complete the tree diagram below, which shows the possible outcomes.
Let E be the event that exactly one four occurs when the pair of dice is thrown.
(b) Calculate P(E). [3]
The pair of dice is now thrown five times.
(c) Calculate the probability that event E occurs exactly three times in the five throws. [3]
(d) Calculate the probability that event E occurs at least three times in the five throws. [3]
Answer/Explanation
Answer:
(a) 
(b) \(P(E)=\frac{1}{6}\times \frac{5}{6}+\frac{5}{6}\times \frac{1}{6}\left ( =\frac{5}{36}+\frac{5}{36} \right )=\frac{10}{36}\left ( =\frac{5}{18} 0r 0.278 \right )\)
(c) \(X\sim B\left ( 5,\frac{5}{18} \right )\)
P(X = 3) = 0.112 [ in fact \(\left ( \frac{5}{3} \right )\left ( \frac{5}{18} \right )^{3}\left ( \frac{13}{18} \right )^{2}=0.112\)
(d) P(X ≥ 3) = 0.135
Question
[Maximum mark: 12]
A bag contains a very large number of ribbons. One quarter of the ribbons are yellow and the rest are blue. Ten ribbons are selected at random from the bag.
(a) Find the expected number of yellow ribbons selected. [2]
(b) Find the probability that exactly six of the ribbons are yellow. [2]
(c) Find the probability that at least two of the ribbons are yellow. [3]
(d) Find the most likely number of yellow ribbons selected. [4]
(e) What assumption have you made about the probability of selecting a yellow ribbon? [1]
Answer/Explanation
Answer:
B(n, p) with n = 10, \(p=\frac{1}{4}\)
(a) \(E(X)=10\times \frac{1}{4}=2.5\)
(b) P(X = 6) = 0.0162
(c) P(X ≥ 2) = 0.756
(d) Since E(X) = 2.5 the mode is 2 or 3
Using GDC
From these values the most likely number of yellow ribbons is 2.
Question
[Maximum mark: 12]
Andrew shoots 20 arrows at a target. He has a probability of 0.3 of hitting the target. All shots are independent of each other. Let X denote the number of arrows hitting the target.
(a) Find the mean and the variance of X. [4]
(b) Find (i) P(X = 5) (ii) P( 4≤ X ≤ 8) [4]
Bill also shoots 20 arrows at a target with probability of 0.3 of hitting the target. All shots are independent of each other.
(c) Calculate the probability that Bill hits the target for the first time on his first shot. [1]
(d) Calculate the probability that Bill hits the target for the first time on his third shot. [3]
Answer/Explanation
Answer:
B(n, p) with n = 20, p = 0.3
(a) Mean = 20×0.3 = 6 Variance = 20×0.3×0.7 = 4.2
(b) (i) P(X = 5) = 0.179 (ii) P( 4≤ X ≤ 8) = 0.780
(c) 0.3
(d) 0.7×0.7×0.3 = 0.147