Question
B and C are subsets of a universal set U such that
\(U = \left\{ {x:x \in \mathbb{Z},0 \leqslant x < 10} \right\},{\text{ }}B = \left\{ {{\text{prime numbers}} < 10} \right\},{\text{ }}C = \left\{ {x:x \in \mathbb{Z},1 < x \leqslant 6} \right\}.\)
List the members of sets
(i) \(B\)
(ii) \(C \cap B\)
(iii) \(B \cup C′\)[4]
Consider the propositions:
p : x is a prime number less than 10.
q : x is a positive integer between 1 and 7.
Write down, in words, the contrapositive of the statement, “If x is a prime number less than 10, then x is a positive integer between 1 and 7.”[2]
Answer/Explanation
Markscheme
(i) \(B = 2, 3, 5, 7\) (A1)
Brackets not required
(ii) \(C \cap B = 2, 3, 5\) (A1)(ft)
Follow through only from incorrect B
(iii) \(C’ = 0, 1, 7, 8, 9\) (A1)(ft)
\(B \cup C’ = 0, 1, 2, 3, 5, 7, 8, 9\) (A1)(ft)
Note: Award (A1) for correct \(C’\) seen. The first (A1)(ft) in (iii) can be awarded only if C was listed incorrectly and a mark was lost as a result in (a)(ii). If C was not listed and \(C’\) is wrong, the first mark is lost. The second mark can (ft) within part (iii) as well as from (i). (C4)[4 marks]
“If x is not a positive integer between 1 and 7, then x is not a prime number less than 10.” (A1)(A1)
Award (A1) for both (not) statements, (A1) for correct order. (C2)[2 marks]
Question
The universal set U is the set of integers from 1 to 20 inclusive.
A and B are subsets of U where:
A is the set of even numbers between 7 and 17.
B is the set of multiples of 3.
List the elements of the following sets:
A,[1]
List the elements of the following sets:
B,[1]
List the elements of the following sets:
\(A \cup B\) ,[2]
List the elements of the following sets:
\(A \cap B’\) .[2]
Answer/Explanation
Markscheme
A = 8, 10, 12, 14, 16 (A1) (C1)[1 mark]
B = 3, 6, 9, 12, 15, 18 (A1) (C1)[1 mark]
\(A \cup B\) = 3, 6, 8, 9,10,12,14,15,16,18 (A2)(ft)
Award (A1) only if a single element is missing or a single extra element is present, (A0) otherwise. (C2)[2 marks]
\(B’\) = 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20 (A1)(ft)
\(A \cap B’\) = 8, 10, 14, 16 (A1)(ft) (C2)[2 marks]
Question
Shade \((A \cup B) \cap C’\) on the diagram below.
[2]
In the Venn diagram below, the number of elements in each region is given.
Find \(n((P \cap Q) \cup R)\).
[2]
\(U\) is the set of positive integers, \({\mathbb{Z}^ + }\).
\(E\) is the set of even numbers.
\(M\) is the set of multiples of \(3\).
(i) List the first six elements of the set \(M\).
(ii) List the first six elements of the set \(E’ \cap M\).[2]
Answer/Explanation
Markscheme
not shading \(C\) or shading \(A \cup B\) (A1)
correct shading (A1) (C2)[2 marks]
Identifying the correct 5 numbers \(3\), \(4\), \(5\), \(6\), \(9\) (A1)
\(27\) (A1) (C2)[2 marks]
(i) \(M = \{ 3{\text{, }}6{\text{, }}9{\text{, }}12{\text{, }}15{\text{, }}18\} \) brackets not required. (A1)
(ii) \(E’ \cap M = \{ 3{\text{, }}9{\text{, }}15{\text{, }}21{\text{, }}27{\text{, }}33\} \) (ft) from (i). (A1)(ft) (C2)[2 marks]
Question
The Venn diagram shows the numbers of pupils in a school according to whether they study the sciences Physics (\(P\)), Chemistry (\(C\)), Biology (\(B\)).
Write down the number of pupils that study Chemistry only.[1]
Write down the number of pupils that study exactly two sciences.[1]
Write down the number of pupils that do not study Physics.[2]
Find \(n[(P \cup B) \cap C]\) .[2]
Answer/Explanation
Markscheme
\(9\) (A1) (C1)[1 mark]
\(12\) (A1) (C1)[1 mark]
\(8 + 3 + 9 + 6\) (M1)
\( = 26\) (A1) (C2)
Note: Award (A1) for \(20\) seen if answer is incorrect.[2 marks]
\(5 + 2 + 3\) (M1)
\( = 10\) (A1) (C2)
Note: Award (A1) for \(29\) or \(19\) seen if answer is incorrect.[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
Consider the universal set \(U = \{ x \in \mathbb{N}|3 < x < 13\} \), and the subsets \(A = \{ {\text{multiples of 3}}\} \) and \(B = \{ 4,{\text{ }}6,{\text{ }}12\} \).
List the elements of the following set.
A[1]
List the elements of the following set.
\(A \cap B’\)[1]
Write down one element of \((A \cup B)’\).[2]
One of the statements in the table below is false. Indicate with an X which statement is false. Give a reason for your answer.
[2]
Answer/Explanation
Markscheme
6, 9, 12 (A1) (C1)[1 mark]
9 (A1)(ft) (C1)
Note: Follow through from their part (a)(i).[1 mark]
any element from {5, 7, 8, 10, 11} (A1)(A1)(ft) (C2)
Note: Award (A1)(ft) for finding \((A \cup B)\), follow through from their A.
Award full marks if all correct elements of \((A \cup B)’\) are listed.[2 marks]
\(15 \notin U\) (R1)(A1) (C2)
Notes: Accept correct reason in words.
If the reason is incorrect, both marks are lost.
Do not award (R0)(A1).[2 marks]
Question
\(U\) is the set of all the positive integers less than or equal to \(12\).
\(A\) , \(B\) and \(C\) are subsets of \(U\) .
\[A = \{ 1{\text{, }}2{\text{, }}3{\text{, }}4{\text{, }}6{\text{, }}12\} \]\[B = \{ {\text{odd integers}}\} \]\[C = \{ 5{\text{, }}6{\text{, }}8\} \]
Write down the number of elements in \(A \cap C\) .[1]
List the elements of \(B\) .[1]
Complete the following Venn diagram with all the elements of \(U\) .
[4]
Answer/Explanation
Markscheme
\(1\) (one) (A1) (C1)
Note: \(6\), \(\{6\} \) or \(\{1\} \) earns no marks.[1 mark]
\(1\), \(3\), \(5\), \(7\), \(9\), \(11\) (A1) (C1)
Note: Do not penalise if braces, parentheses or brackets are seen.[1 mark]
(A1)(A1)(ft)(A1)(ft)(A1)(ft) (C4)
Notes: Award (A1) for the empty set \(A \cap B \cap C\) .
Award (A1)(ft) for the correct placement of \(6\), \(5\), \(1\) and \(3\).
Award (A1)(ft) for the correct placement of \(2\), \(4\), \(12\), \(7\), \(9\), \(11\), \(8\).
Award (A1)(ft) for the correct placement of \(10\).
Follow through from part (b).[4 marks]
Question
A group of 33 people was asked about the passports they have. 21 have Australian passports, 15 have British passports and 3 have neither.
A group of 33 people was asked about the passports they have. 21 have Australian passports, 15 have British passports and 3 have neither.
Find the number that have both Australian and British passports.[2]
In the Venn diagram below, set A represents the people in the group with Australian passports and set B those with British passports.
Write down the value of
(i) q ;
(ii) p and of r .[2]
In the Venn diagram below, set A represents the people in the group with Australian passports and set B those with British passports.
Find \(n(A \cup B’)\) .[2]
Answer/Explanation
Markscheme
\(21 + 15 + 3 – 33\) or equivalent (M1)
Note: Award (M1) for correct use of all four numbers.
\( = 6\) (A1) (C2)[2 marks]
(i) q = 6 (A1)(ft)
(ii) p =15, r = 9 (A1)(ft) (C2)
Note: Follow through from their answer to part (a).[2 marks]
15 + 6 + 3 (M1)
Note: Award (M1) for their figures seen in a correct calculation:
15 + 6 + 3 or 21 + 3 or 33 − 9
= 24 (A1)(ft) (C2)
Note: Follow through from parts (a) and (b) or from values shown on Venn diagram.[2 marks]
Question
Music lessons in Piano (P), Violin (V) and Flute (F) are offered to students at a school.
The Venn diagram shows the number of students who learn each kind of instrument.
Write down the total number of students in the school.[1]
Write down the number of students who
(i) learn violin only;
(ii) learn piano or flute or both;
(iii) do not learn flute.[3]
Explain, in words, the meaning of the part of the diagram that represents the set \(P \cap F’\).[2]
Answer/Explanation
Markscheme
145 (A1) (C1)[1 mark]
(i) 56 (A1)
(ii) 85 (A1)
(iii) 89 (A1) (C3)[3 marks]
The students who learn the piano and do not learn the flute. (A1)(A1) (C2)
Notes: Award (A1) for students who learn piano, not flute, (A1) for and (accept but). Accept correct alternative statements. Accept “The number of students who learn the piano and do not learn the flute”.[2 marks]
Question
\(U = \{ x|x{\text{ is an integer, }}2 < x < 10\}\)
A and B are subsets of U such that A = {multiples of 3}, B = {factors of 24}.
List the elements of
(i) U ;
(ii) B .[2]
Write down the elements of U on the Venn diagram.
[3]
Write down \(n (A \cap B)\).[1]
Answer/Explanation
Markscheme
(i) 3, 4, 5, 6, 7, 8, 9 (A1)
(ii) 3, 4, 6, 8 (A1)(ft) (C2)
Notes: Follow through from part (a)(i).[2 marks]
(A1)(ft) for their 3, 6
(A1)(ft) for their 4, 8, 9
(A1)(ft) for their 5, 7 (A1)(ft)(A1)(ft)(A1)(ft) (C3)
Note: Follow through from their universal set and set B in part (a).[3 marks]
2 (A1)(ft) (C1)
Note: Follow through from their Venn diagram.[1 mark]
Question
\(U\) is the set of positive integers less than or equal to \(10\).
\(A\), \(B\) and \(C\) are subsets of \(U\).
\(A = \left\{ {{\text{even integers}}} \right\}\)
\(B = \left\{ {{\text{multiples of }}3} \right\}\)
\(C = \left\{ {6,{\text{ }}7,{\text{ }}8,{\text{ }}9} \right\}\)
List the elements of \(A\).[1]
List the elements of \(B\).[1]
Complete the Venn diagram with all the elements of \(U\).
[4]
13N.1.sl.TZ0.2
Markscheme
\(2, 4, 6, 8, 10\) (A1) (C1)
Note: Do not penalize the use of \(\left\{ {{\text{ }}} \right\}\).[1 mark]
\(3, 6, 9\) (A1) (C1)
Note: Do not penalize the use of \(\left\{ {{\text{ }}} \right\}\).
Follow through from part (a) only if their \({\text{U}}\) is listed.[1 mark]
(A1)(ft)(A1)(ft)(A1)(ft)(A1)(ft) (C4)
Notes: Award (A1)(ft) for the correct placement of \(6\).
Award (A1)(ft) for the correct placement of \(8\) and \(9\) and the empty region.
Award (A1)(ft) for the correct placement of \(2\), \(4\), \(3\), \(7\), and \(10\).
Award (A1)(ft) for the correct placement of \(1\) and \(5\).
If an element is in more than one region, award (A0) for that element.
Follow through from their answers to parts (a) and (b).[4 marks]
Question
Tuti has the following polygons to classify: rectangle (R), rhombus (H), isosceles triangle (I), regular pentagon (P), and scalene triangle (T).
In the Venn diagram below, set \(A\) consists of the polygons that have at least one pair of parallel sides, and set \(B\) consists of the polygons that have at least one pair of equal sides.
Complete the Venn diagram by placing the letter corresponding to each polygon in the appropriate region. For example, R has already been placed, and represents the rectangle.[3]
State which polygons from Tuti’s list are elements of
(i) \(A \cap B\);
(ii) \((A \cup B)’\).[3]
Answer/Explanation
Markscheme
(A3) (C3)
Note: Award (A3) if all four letters placed correctly,
(A2) if three letters are placed correctly,
(A1) if two letters are placed correctly.
(i) Rhombus and rectangle OR H and R (A1)(ft)
(ii) Scalene triangle OR T (A2)(ft) (C3)
Notes: Award (A1) for a list R, H, I, P seen (identifying the union).
Follow through from their part (a).
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
Consider the following Venn diagrams. Each diagram is shaded differently.
In the following table there are six sets. Each of these sets corresponds to the shaded region of one of the Venn diagrams. In the correct space, write the number of the diagram that corresponds to that set.
Answer/Explanation
Markscheme
(A6)(C6)
Note: Award (A1) for each correct entry.
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
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
Consider the following Venn diagrams.
Write down an expression, in set notation, for the shaded region represented by Diagram 1.[1]
Write down an expression, in set notation, for the shaded region represented by Diagram 2.[1]
Write down an expression, in set notation, for the shaded region represented by Diagram 3.[2]
Shade, on the Venn diagram, the region represented by the set \(\left( {H \cup I} \right)’\).
[1]
Shade, on the Venn diagram, the region represented by the set \(J \cap K\).
[1]
Answer/Explanation
Markscheme
A’ (A1)
Note: Accept alternative set notation for complement such as U − A.[1 mark]
\(C \cap D’\) OR \(D’ \cap C\) (A1)
Note: Accept alternative set notation for complement.[1 mark]
\(\left( {E \cap F} \right) \cup G\) OR \(G \cup \left( {E \cap F} \right)\) (A2) (C4)
Note: Accept equivalent answers, for example \(\left( {E \cup G} \right) \cap \left( {F \cup G} \right)\).[2 marks]
(A1)[1 mark]
(A1) (C2)[1 mark]
Question
The table below shows the number of words in the extended essays of an IB class.
Draw a histogram on the grid below for the data in this table.
[3]
Write down the modal group.[1]
The maximum word count is \(4000\) words.
Write down the probability that a student chosen at random is on or over the word count.[2]
Answer/Explanation
Markscheme
(A3) (C3)
Notes: (A3) for correct histogram, (A2) for one error, (A1) for two errors, (A0) for more than two errors.
Award maximum (A2) if lines do not appear to be drawn with a ruler.
Award maximum (A2) if a frequency polygon is drawn.[3 marks]
\({\text{Modal group}} = 3800 \leqslant w < 4000\) (A1) (C1)[1 mark]
\({\text{Probability}} = \frac{3}{{35}}{\text{ }}(0.0857{\text{, }}8.57\% )\) (A1)(A1) (C2)
Note: (A1) for correct numerator (A1) for correct denominator.[2 marks]
Question
The following histogram shows the weights of a number of frozen chickens in a supermarket. The weights are grouped such that \(1 \leqslant {\text{weight}} < 2\), \(2 \leqslant {\text{weight}} < 3\) and so on.
Find the total number of chickens.[1]
Write down the modal group.[1]
Gabriel chooses a chicken at random.
Find the probability that this chicken weighs less than \(4{\text{ kg}}\).[2]
Answer/Explanation
Markscheme
\(96\) (A1) (C1)[1 mark]
\(3 \leqslant {\text{weight}} < 4{\text{ kg}}\) . Accept \(3 – 4{\text{ kg}}\) (A1) (C1)[1 mark]
For adding three heights or subtracting \(14\) from \(96\) (M1)
\(\frac{{82}}{{96}}{\text{ }}(0.854{\text{ or }}\frac{{41}}{{48}}{\text{, }}85.4\% )\) (ft) from (b). (A1)(ft) (C2)[2 marks]
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
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
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
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
The IB grades attained by a group of students are listed as follows.
\[{\text{6}}\;\;\;{\text{4}}\;\;\;{\text{5}}\;\;\;{\text{3}}\;\;\;{\text{7}}\;\;\;{\text{3}}\;\;\;{\text{5}}\;\;\;{\text{4}}\;\;\;{\text{2}}\;\;\;{\text{5}}\]
Find the median grade.[2]
Calculate the interquartile range.[2]
Find the probability that a student chosen at random from the group scored at least a grade \(4\).[2]
Answer/Explanation
Markscheme
\(2\;\;\;3\;\;\;3\;\;\;4\;\;\;4\;\;\;5\;\;\;5\;\;\;5\;\;\;6\;\;\;7\) (M1)
Note: Award (M1) for correct ordered set.
\(({\text{Median}} = ){\text{ }}4.5\) (A1) (C2)
\(5 – 3\) (M1)
Note: Award (M1) for correct quartiles seen.
\( = 2\) (A1) (C2)
\(\frac{7}{{10}}\;\;\;(0.7,{\text{ }}70\% )\) (A2) (C2)
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
The probability that Nikita wins a tennis match depends on the surface of the tennis court on which she is playing. The probability that she plays on a grass court is \(0.4\). The probability that Nikita wins on a grass court is \(0.35\). The probability that Nikita wins when the court is not grass is \(0.25\).
Complete the following tree diagram.
Find the probability that Nikita wins a match.[3]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct entry.[3 marks]
\(0.4 \times 0.35 + 0.6 \times 0.25\) (A1)(ft)(M1)
Note: Award (A1)(ft) for two correct products from their tree diagram seen, and (M1) for the addition of their products.
\( = 0.29\) (A1)(ft) (C3)
Note: Follow through from part (a).[3 marks]
Question
Two friends, Sensen and Cruz, are conducting an investigation on probability.
Sensen has a fair six-sided die with faces numbered \(1,\,\,2,\,\,2,\,\,4,\,\,4\) and \(4\). Cruz has a fair disc with one red side and one blue side.
The die and the disc are thrown at the same time.
Find the probability that the number shown on the die is \(1\) and the colour shown on the disc is blue;[2]
Find the probability that the number shown on the die is \(1\) or the colour shown on the disc is blue;[2]
Find the probability that the number shown on the die is even given that the colour shown on the disc is red.[2]
Answer/Explanation
Markscheme
\(\frac{1}{2} \times \frac{1}{6}\) (M1)
\(\frac{1}{{12}}\,\,(0.0833,\,\,8.33\,\% ,\,\,0.08333…)\) (A1) (C2)
\(\frac{1}{2} + \left( {\frac{1}{2} \times \frac{1}{6}} \right)\) (M1)
OR
\(\frac{1}{6} + \frac{1}{2} – \frac{1}{{12}}\,\) (M1)
\(\frac{7}{{12}}\,\,(0.583,\,\,58.3\,\% ,\,\,0.58333…)\) (A1) (C2)
Note: Award (M1)(A0) for a correct attempt at a possibility/sample space diagram or tree diagram or \(\frac{1}{6} + \left( {\frac{5}{6} \times \frac{1}{2}} \right)\), leading to an incorrect answer.
\(\frac{1}{3} + \frac{1}{2}\) (M1)
OR
\(\frac{{\frac{5}{6} \times \frac{1}{2}}}{{\frac{1}{2}}}\) (M1)
\(\frac{5}{6}\,\,(0.833,\,\,83.3\,\% ,\,\,0.83333…)\) (A1) (C2)
Notes: Award (M1)(A0) for a correct attempt at a possibility/sample space diagram or tree diagram, leading to an incorrect answer.
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]
[MAI 4.5-4.7] PROBABILITY I (VENN DIAGRAMS – TABLES)-neha
Question
The following Venn diagram shows the sample space \(U\) and the event \(A\) and \(B\) together with the numbers of elements in the corresponding regions.
(a) Complete the following table.
(b) Write down the following probabilities
(c) Write down the following conditional probabilities
Answer/Explanation
Ans
(a) 
(b) 
(c) 
Question
The following Venn diagram shows the sample space \(U\) and the events \(A\) and \(B\) together with their probabilities in the corresponding regions.
(a) Write down the following probabilities
(b) Write down the following conditional probabilities
Answer/Explanation
Ans
(a) 
(b) 
Question
The following table shows the distribution of a population according to two criteria, gender and group. We select a person at random.
(a) Write down the following probabilities
(b) Write down the following conditional probabilities
Answer/Explanation
Ans
(a) 
(b) 
Question
The following Venn diagram shows the universal set \(U\) and the sets \(A\) and \(B\) together with the probabilities of the corresponding regions.
where \(a + b + c + d = 1\).
(a) Express in terms of \(a, b, c, d\) the following probabilities:
(b) Express in terms of \(a, b, c, d\) the following probabilities:
(c) Given that the events \(A\) and \(B\) are independent and \(a = b = c\) , find the values of \(a, b, c\) and \(d\)
Answer/Explanation
Ans
Question
The following Venn diagram shows the universal set \(U\) and the sets \(A\) and \(B\) together with the probabilities of the corresponding regions.
(a) Write down the values of \(P(A)\) and \(P(A \cup B)\)
(b) Write down the value of \(x\) given that \(A\) and \(B\) are mutually exclusive.
(c) Find the value of \(x\) given that \(A\) and \(B\) are independent.
(d) Find the value of \(x\) given that \(P(A | B) = 0.5\).
(e) Find the value of \(x\) given that \(P(B | A) = 0.25\).
Answer/Explanation
Ans
(a) \(P(A) = 0.4, P(A \cup B) = 0.7\)
(b) \(x = 0\)
(c) \(x = 2.0\)
(d) \(x = 3.0\)
(e) \(x = 1.0\)
Question
The following Venn diagram shows the universal set \(U\) and the sets \(A\) and \(B\).
(a) Find \(n(B \cap A’)\)
(b) An element is selected at random from \(U\). What is the probability that this element is in \(P(B \cap A’)\)?
Answer/Explanation
Ans
(a) \(n(A \cup B) = n(A) + n(B) – n(A \cap B)\)
\(65 = 30 + 50 – n(A \cap B) \Rightarrow n(A \cap B) = 15\) (may be on the diagram)
\(n(B \cap A’) = 50 – 15 = 35\)
(b) \(P(B \cap A’)= \frac{n(B \cap A’)}{n(U)}=\frac{35}{100}=0.35\)
Question
The following Venn diagram shows a sample space \(U\) and events \(A\) and \(B\).
(a) Find (i) \(n(A \cap B)\); (ii) \(P(A \cap B)\)
(b) Explain why events \(A\) and \(B\) are not mutually exclusive.
Answer/Explanation
Ans
(a) (i) \(n(A \cap B) = 2\)
(ii) \(P(A \cap B) = \frac{2}{36}\left ( or \frac{1}{18} \right )\)
(b) \(n(A \cap B)\neq 0\) (or equivalent)
Question
The Venn diagram below shows events \(A\) and \(B\) where \(P(A) = 0.3, P(A\cup B)=0.6\) and \(P(A \cap B)=0.1\). The values \(m, n, p\) and \(q\) are probabilities.
(a) (i) Write down the value of \(n\).
(ii) Find the value of \(m\) , of \(p\) , and of \(q\).
(b) Find \(P( B’)\) .
Answer/Explanation
Ans
(a) (i) \(n = 0.1\)
(ii) \(m = 0.2, p = 0.3, q = 0.4\)
(b) \(P(B′) = 0.6\)
Question
[without GDC]
Consider the events \(A\) and \(B\) , where \(P(A)=0.5, P(B)=0.7\) and \(P(A \cap B)=0.3\).
The Venn diagram below shows the events \(A\) and \(B\) , and the probabilities \(p\) , \(q\) , \(r\) .
(a) Write down the value of (i) \(p\) ; (ii) \(q\) ; (iii) \(r\) .
(b) Find the value of \(P(A | B’)\).
(c) Hence, or otherwise, show that the events \(A\) and \(B\) are not independent.
Answer/Explanation
Ans
(a) (i) \(p = 0.2\)
(ii) \(q = 0.4\)
(iii) \(r = 0.1\)
(b) \(P(A | B’)=\frac{2}{3}\)
(c) valid reason e.g. \(\frac{2}{3} \neq 0.5, 0.35 \neq 0.3\) thus, \(A\) and \(B\) are not independent
Question
For the events \(A\) and \(B\) , \(P(A)=0.6, P(B)=0.8\) and \(P(A \cup B)=1\).
(a) Find \(P(A \cap B)\);
(b) Find \(P(A’ \cup B’)\)
Answer/Explanation
Ans
(a) \(p(A \cap B) = 0.6 + 0.8 – 1 = 0.4\)
(b) \(p(A’ \cup B’) = p((A \cap B) ‘) = 1 – 0.4 = 0.6\)
Question
The Venn diagram below shows information about \(120\) students in a school. Of these, \(40\) study Chinese \((C)\), \(35\) study Japanese \((J)\), and \(30\) study Spanish \((S)\).
A student is chosen at random from the group. Find the probability that the student
(a) studies exactly two of these languages;
(b) studies only Japanese;
(c) does not study any of these languages.
Answer/Explanation
Ans
(a) \(\frac{19}{120} \left ( = 0.158 \right )\)
(b) \(35 – (8 + 5 + 7)(= 15)\)
Probability = \(\frac{15}{120} \left ( = \frac{3}{24} = \frac{1}{8} = 0.125\right )\)
(c) Number studying = \(76\)
Number not studying = \(120\) – number studying = \(44\)
Probability = \(\frac{44}{120} \left ( =\frac{11}{30} = 0.367 \right )\)
Question
In a class, \(40\) students take chemistry only, \(30\) take physics only, \(20\) take both chemistry and physics, and \(60\) take neither.
(a) Find the probability
(i) that a student takes physics given that the student takes chemistry.
(ii) that a student takes physics given that the student does not take chemistry.
(b) State whether the events “taking chemistry” and “taking physics” are mutually exclusive, independent, or neither. Justify your answer.
Answer/Explanation
Ans
(a) (i) \(P(P | C) = \frac{20}{20 + 40} = \frac{1}{3}\)
(ii) \(P(P | C) = \frac{30}{30 + 60} = \frac{1}{3}\)
(b) Investigating conditions, or some relevant calculations \(P\) is independent of \(C\), with valid reason
Question
In a survey, \(100\) students were asked “do you prefer to watch television or play sport?”
Of the \(46\) boys, \(33\) said they would choose sport, while \(29\) girls made this choice.
By completing this table or otherwise, find the probability that
(a) a student selected at random prefers to watch television;
(b) a student prefers to watch television, given that the student is a boy.
Answer/Explanation
Ans
(a) 
\(P(TV) = \frac{38}{100}\)
(b) \(P(TV | Boy) = \frac{13}{46}\)
Question
In a survey of \(200\) people, \(90\) of whom were female, it was found that \(60\) people were unemployed, including \(20\) males.
(a) Using this information, complete the table below.
(b) If a person is selected at random from this group of \(200\), find the probability that this person is
(i) an unemployed female;
(ii) a male, given that the person is employed.
Answer/Explanation
Ans
(a) 
(b) (i) \(P\)(unemployed female) = \(\frac{40}{200} =\frac{1}{5}\)
(ii) \(P\)(male I employed person) = \(\frac{90}{140} =\frac{9}{14}\)
Question
The eye colour of \(97\) students is recorded in the chart below.
One student is selected at random. Write down
(a) the probability that the student is a male.
(b) the probability that the student has green eyes, given that the student is a female.
(c) Find the probability that the student has green eyes or is male.
Answer/Explanation
Ans
(a) \(46/97\)
(b) \(13/51\)
(c) \(59/97\)
Question
There are \(20\) students in a classroom. Each student plays only one sport. The table below gives their sport and gender.
(a) One student is selected at random.
(i) Calculate the probability that the student is a male or is a tennis player.
(ii) Given that the student selected is female, calculate the probability that the student does not play football.
(b) Two students are selected at random. Calculate the probability that neither student plays football.
Answer/Explanation
Ans
(a) (i) P(male or tennis) = \(\frac{12}{20} \left ( = \frac{3}{5} \right )\)
(ii) P(not football | female) = \(\frac{6}{11}\)
(b) P(first not football) = \(\frac{6}{11}\), P(second not football) = \(\frac{10}{19}\)
P(neither football) = \(\frac{11}{20} \times \frac{10}{19}\)
P(neither football) = \(\frac{110}{380} \left ( = \frac{11}{38} \right )\)
Question
Consider events \(A\), \(B\) such that \(P(A)\neq 0, P(A)\neq 1, P(B)\neq 0\) and \(P(A)\neq 1\)
In each of the situations (i), (ii), (iii) below state whether \(A\) and \(B\) are
mutually exclusive (M); independent (I); neither (N).
(i) \(P(A | B)= P(A)\) (ii) \(P(A \cap B) = 0\) (iii) \(P(A \cap B) = P(A)\)
Answer/Explanation
Ans
(a) Independent (I)
(b) Mutually exclusive (M)
(c) Neither (N)
Question
Consider the independent events \(A\) and \(B\).
Given that \(P(B) = 2P(A)\) and \(P(A \cup B)\), find \(P(B)\).
[Hint: For independent events it also holds: \( P(A \cup B) = P(A)+P(B)-P(A)P(B)\)]
Answer/Explanation
Ans
for independence \(P(A \cap B) = P(A)\times P(B)\)
\(P(A \cup B) = P(A)+P(A)-P(A)\times P(B)\)
\(0.52 = P(A)+2P(A)-2P(A)P(A)\)
\(P(B)=0.4\)
Question
The letters of the word PROBABILITY are written on \(11\) cards as shown below.
Two cards are drawn at random without replacement.
Let A be the event the first card drawn is the letter A.
Let B be the event the second card drawn is the letter B.
(a) Find \(P( A)\).
(b) Find \(P( B | A )\).
(c) Find \(P(A \cap B)\).
Answer/Explanation
Ans
(a) \(P(A) = \frac{1}{11}\)
(b) \(P(B│A) = \frac{2}{10}\)
(c) \(P(A ∩ B) = \frac{1}{11} \times \frac{2}{10} = \frac{2}{110}\)
Question
The following diagram shows a circle divided into three sectors \(A\), \(B\) and \(C\). The angles at the centre of the circle are \(90°\), \(120°\) and \(150°\). Sectors \(A\) and \(B\) are shaded
as shown. The arrow is spun. It cannot land on the lines between the sectors. Let \(A, B, C\) and \(S\) be the events defined by
(a) Find \(P(B)\) ;
(b) Find \(P(S)\);
(c) Find \(P( A| S)\).
Answer/Explanation
Ans
(a) \(\frac{120}{360} \left ( = \frac{1}{3}=0.333 \right )\)
(b) \(\frac{90+120}{360} \left ( = \frac{210}{360}=\frac{7}{12}=0.583 \right )\)
(c) \(\frac{90}{210} \left ( = \frac{3}{7}=0.429 \right )\)
Question
Let \(A\) and \(B\) be events such that \(P(A)=\frac{1}{2}, P(A)=\frac{3}{4}\) and \(P(A \cup B)=\frac{7}{8}\).
(a) Calculate \(P(A \cap B)\).
(b) Calculate \(P(A | B)\).
(c) Are the events \(A\) and \(B\) independent? Give a reason for your answer.
Answer/Explanation
Ans
(a) \(P(A \cup B) = P(A)+P(B)-P(A \cap B)\)
\(P(A \cap B) = \frac{1}{2}+\frac{3}{4}-\frac{7}{8}=\frac{3}{8}\)
(b) \(P(A | B) = \frac{P(A \cap B)}{P(B)}\left ( =\frac{\frac{3}{8}}{\frac{3}{4}} \right )=\frac{1}{2}\)
(c) Yes, the events are independent
EITHER \(P(A | B) =P(A)\) OR \(P(A \cap B) = P(A)P(B)\)
Question
Let \(A\) and \(B\) be independent events such that \(P(A ) = 0.3\) and \(P( B) = 0.8\).
(a) Find \(P(A \cap B)\).
(b) Find \(P(A \cup B)\).
(c) Are \(A\) and \(B\) mutually exclusive? Justify your answer.
Answer/Explanation
Ans
(a) Independent \(\Rightarrow P(A \cap B)= P(A)\times P(B)= 0.3\times 0.8=0.24\)
(b) \(P(A \cup B)=P(A) + P(B)-P(A \cap B) = 0.3+0.8-0.24=0.86\)
(c) No, since \(P(A \cap B)\neq 0\) or \(P(A \cup B)\neq P(A) + P(B)\)
Question
Events \(E\) and \(F\) are independent, with \(P(E)=\frac{2}{3}\) and \(P(E\cap F)=\frac{1}{3}\). Calculate
(a) \(P(F)\) ;
(b) \(P(E\cup F\).
Answer/Explanation
Ans
(a) \(P(F)= \frac{P(E \cap F)}{P(E)}, P(F)= \frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}\)
(b) \(P(E \cup F) = P(E)+P(F)-P(E \cap F) = \frac{2}{3}+\frac{1}{2}-\frac{1}{3}=\frac{5}{6}(=0.833)\)
Question
Consider the events A and B , where \(P(A)=\frac{2}{5}\), \(P(B’)=\frac{1}{4}\) and \(P(A\cap B)=\frac{7}{8}\).
(a) Write down \(P(B)\).
(b) Find \(P(A \cap B)\).
(c) Find \(P(A | B)\).
Answer/Explanation
Ans
(a) \(\frac{3}{4}\)
(b) \(P(A\cup B)= P(A)+P(B)-P(A\cap B)\)
\(P(A\cap B)= P(A)+P(B)-P(A\cup B)= \frac{2}{5}+\frac{3}{4}-\frac{7}{8}=\frac{11}{40}(0.275)\)
(c) \(P(A | B)=\frac{P(A\cap B)}{P(B)} \left ( =\frac{\frac{11}{40}}{\frac{3}{4}} \right )= \frac{11}{30} (0.367)\)
Question
Let \(A\) and \(B\) be independent events, where \(P( A) = 0.6\) and \(P( B) = x\) .
(a) Write down an expression for \(P(A \cap B)\).
(b) Given that \(P(A \cup B) = 0.8\)
(i) find \(x\); (ii) find \(P(A \cap B)\).
(c) Hence, explain why \(A\) and \(B\) are not mutually exclusive.
Answer/Explanation
Ans
(a) \(P(A \cap B) = P(A) \times P(B) = 0.6x\)
(b) (i) \(P(A \cup B) = P(A) + P(B) – P(A)P(B)\)
\(\Leftrightarrow 0.80 = 0.6 + x – 0.6x \Leftrightarrow 0.2 = 0.4x \Leftrightarrow x = 0.5\)
(ii) \(P(A \cap B) = 0.3\)
(c) \(P(A \cap B) \neq 0\)
Question
The events \(A\) and \(B\) are independent such that and \(P(B) = 3P(A)\) and \(P(A\cup B)=0.68\). Find \(P(B)\).
Answer/Explanation
Ans
Let \((A) = x\) then \(P(B) = 3x\) and \(P(A \cap B) = P(A) \times 3P(A) = 3x^{2}\)
\(P(A \cup B) = P(A) + P(B) – P(A \cap B) \Leftrightarrow 0.68 = x + 3x – 3x^{2}\)
\(3x^{2} – 4x + 0.68 = 0 \Leftrightarrow x = 0.2\) (\(x = 1.133\), not possible)
\(P(B) = 3x = 0.6\)
Question
For events \(A\) and \(B\) , the probabilities are \(P(A)=\frac{3}{11}\), \(P(B)=\frac{4}{11}\).
Calculate the value of \(P(A \cap B)\).
(a) if \(P(A \cup B)=\frac{6}{11}\);
(b) if events \(A\) and \(B\) are independent.
Answer/Explanation
Ans
(a) \(P (A \cap B) = P (A) + P (B) – P (A \cup B) = \frac{3}{11}+\frac{4}{11}-\frac{6}{11}=\frac{1}{11} (0.0909)\)
(b) For independent events, \(P (A \cap B) = P (A) \times P (B) = \frac{3}{11}\times \frac{4}{11}= \frac{12}{121} (0.0992)\)
Question
The events \(A\) and \(B\) are such that \(P(A)=0.5\), \(P(B)=0.3\) and \(P(A\cup B)=0.6\)
(a) (i) Find the value of \(P(A\cap B)\)
(ii) Hence show that \(A\) and are not independent.
(b) Find the value of \(P(B | A)\) .
Answer/Explanation
Ans
(a) (i) Use of \(P(A \cup B) = P(A)+P(B)-P(A\cap B)\)
\(0.6 = 0.5+0.3-P(A\cap B)\)
\(P(A\cap B) = 0.2\)
(ii) P(A)P(B)=0.15\neq P(A\cap B)
Hence not independent
(b) Use of \(P(B|A) = \frac{P(B\cap A)}{P(A)}\)
=\(\frac{0.2}{0.5}\)
=\(0.4\)
Question
Let \(A\) and \(B\) be events such that \(P(A)=0.6, P(A\cup B)=0.8\) and \(P(B | A )= 0.6\). Find \(P(B )\).
Answer/Explanation
Ans
As \(P(A | B) = P(A)\) then \(A\) and \(B\) are independent events
Using \(P(A \cup B) = P(A) + P(B) – P(A) \times P(B)\)
to obtain \(0.8 = 0.6 + P(B) – 0.6 \times P(B)\)
\(0.8 = 0.6 + 0.4 P(B) \Rightarrow P(B) = 0.5 \) OR by using Venn diagram
Question
Let \(A\) and \(B\) be events such that \(P(A)=\frac{1}{5}\), \(P(A | B)=\frac{1}{4}\) and \(P(A\cup B)=\frac{7}{10}\).
(a) Find \(P(A\cap B)\).
(b) Find \(\(P(B)\).
(c) Show that \(A\) and \(B\) are not independent.
Answer/Explanation
Ans
(a) \(P(A\cap B)=P(A)\times P(B|A)\)
=\(\frac{1}{5}\times \frac{1}{4} \left ( =\frac{1}{20} \right )\)
(b) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(\Rightarrow P(B)=\frac{7}{10}-\frac{1}{5}+\frac{1}{20}\)
=\(\frac{11}{20}\)
(c) METHOD 1
\(P(B)=\frac{11}{20} and P(B|A)=\frac{1}{4}\)
\(P(B)\neq P(B|A)\)
\(\Rightarrow\) \(A\) and \(B\) are not independent
METHOD 2
\(P(A) \times P(B) =\frac{1}{5} \times and \frac{11}{20} and P(A \cap B)=\frac{1}{20}\)
\(P(A \cap B) \neq P(A) \times P(B)\)
\(\Rightarrow\) \(A\) and \(B\) are not independent
Question
Given that events \(A\) and \(B\) are independent with \(P(A\cap B)=0.3, P(A\cap B’)=0.3\), find \(P(A\cup B)\).
Answer/Explanation
Ans
(Venn diagram)
\(P(A \cap B) = P(A)P(B) 0.3 = 0.6 \times P(B)\)
\(P(B) = 0.5\)
Therefore, \(P(A \cup B) = 0.8\)
Question
The independent events \(A\) and \(B\) are such that \(P(A)=0.4\) and \(P(A\cup B)=0.88\). Find
(a) \(P (B )\).
(b) the probability that either \(A\) occurs or \(B\) occurs, but not both.
Answer/Explanation
Ans
(a) \(0.88 = 0.4 + P(B) – 0.4P(B)\)
\(0.6P(B) = 0.48 = > P(B) = 0.8\)
(b) \(P(A \cup B) – P(A \cap B) = 0.88 – 0.32= 0.56\)
Question
In a bilingual school there is a class of \(21\) pupils. In this class, \(15\) of the pupils speak Spanish as their first language and \(12\) of these \(15\) pupils are Argentine. The other \(6\) pupils
in the class speak English as their first language and \(3\) of these \(6\) pupils are Argentine. A pupil is selected at random from the class and is found to be Argentine. Find the probability that
the pupil speaks Spanish as his/her first language.
Answer/Explanation
Ans
Using a tree diagram,
\(p (S|A) = \frac{12}{15}=\frac{4}{5}\)
Question
Two unbiased \(6\)-sided dice are rolled, a red one and a black one. Let \(E\) and \(F\) be the events
\(E\) : the same number appears on both dice;
\(F\) : the sum of the numbers is \(10\).
(a) Find \(P( E )\).
(b) Find \(P( F )\).
(c) Find \(P(E\cup F)\).
Answer/Explanation
Ans
Total number of possible outcomes = \(36\)
(a) \(P(E) = P(1,1)+P(2,2)+P(3,3)+P(4,4)+P(5,5)+P(6,6)=\frac{6}{36}\)
(b) \(P(F) = P(6,4)+P(5,5)+P(4,6)=\frac{3}{36}\)
(c) \(P(E\cap F)=P(E)+P(F)-P(E\cap F)\)
\(P(E\cap F)=\frac{1}{36}\)
\(P(E\cap F)=\frac{6}{36}+\frac{3}{36}-\frac{1}{36}\left ( \frac{8}{36}=\frac{2}{9}, 0.222 \right )\)
Question
Two fair dice are thrown and the number showing on each is noted. The sum of these two numbers is \(S\). Find the probability that
(a) \(S\) is less than \(8\);
(b) at least one die shows a \(3\);
(c) at least one die shows a \(3\), given that \(S\) is less than \(8\).
Answer/Explanation
Ans
Sample space ={(1, 1), (1, 2) … (6, 5), (6, 6)}
(This may be indicated in other ways, e.g, a grid or a tree diagram, partly or fully completed)
(a) \(P (S < 8) = 21/36 = 7/12\)
(b) P (at least one \(3\)) = \(\frac{11}{36}\)
(c) P (at least one \(3|S < 8) = 7/21 = \frac{1}{3}\)
Question
In a survey of \(50\) people it is found that \(40\) own a television and \(21\) own a computer. Four do not own either a computer or a television. A person is chosen at random from this group.
Find the probability
(a) that this person owns both a television and a computer.
(b) that this person owns a television, given that he also owns a computer.
(c) that this person owns both a television or a computer.
(d) that this person owns a television, given that he does not own a computer.
(e) that this person owns a computer, given that he does not own a television.
Answer/Explanation
Ans
(a) METHOD 1
\(P(T\cap C)=0.3\)
METHOD 2
\(n(T\cup C)=n(T)+n(C)-n(T\cap C)\)
\(46=40+21-n(T\cap C)\)
\(n(T\cap C)=15\)
\(P(T\cap C)=0.3\)
(b) \(P(T|C)=\frac{P(T\cap C)}{P(C)}\)
=\(\frac{0.3}{0.42} (=\frac{5}{7}, 0.714 to 3\) s.f.)
Question
In a school of \(88\) boys, \(32\) study economics (E), \(28\) study history (H) and \(39\) do not study either subject. This information is represented in the following Venn diagram.
(a) Calculate the values \(a, b, c\) .
(b) A student is selected at random.
(i) Calculate the probability that he studies both economics and history.
(ii) Given that he studies economics, calculate the probability that he does not study history.
(c) A group of three students is selected at random from the school.
(i) Calculate the probability that none of these students studies economics.
(ii) Calculate the probability that at least one of these students studies economics.
Answer/Explanation
Ans
(a) 
\(n (E \cup H) = a + b + c = 88 – 39 = 49\)
\(n (E \cup H) = 32 + 28 – b = 49 \Leftrightarrow b = 11\)
\(a = 32 – 11 = 21\)
\(c = 28 – 11 = 17\)
(b) (i) \(P(E\cap H)=\frac{11}{88}=\frac{1}{8}\)
(ii) \(P(H’|E)=\frac{P(H’\cap E)}{P(E)}=\frac{\frac{21}{88}}{\frac{32}{88}}=\frac{21}{32}(=0.656) Or directly =\frac{21}{32}\)
(c) (i) P(none in economics) = \(\frac{56\times 55\times 54}{88\times 87\times 86}=0.253\)
(ii) P(at least one) = \(1-0.253 = 0.747\)
OR
\(3\left ( \frac{32}{88}\times \frac{56}{87}\times \frac{55}{86} \right )+3\left ( \frac{32}{88}\times \frac{31}{87}\times \frac{56}{86} \right )+\frac{32}{88}\times \frac{31}{87}\times \frac{30}{86}=0.747 \)
Question
Two restaurants, Center and New, sell fish rolls and salads.
Let \(F\) be the event a customer chooses a fish roll.
Let \(S\) be the event a customer chooses a salad.
Let \(N\) be the event a customer chooses neither a fish roll nor a salad.
In the Center restaurant \(P(F)=0.31, P(S)= 0.62, P(N )=0.14\).
(a) Show that \(P(F\cup S)=0.07\).
(b) Given that a customer chooses a salad, find the probability the customer also chooses a fish roll.
(c) Are \(F\) and \(S\) independent events? Justify your answer.
At New restaurant, \(P(N )= 0.14\). Twice as many customers choose a salad as choose a fish roll. Choosing a fish roll is independent of choosing a salad.
(d) Find the probability that a fish roll is chosen.
Answer/Explanation
Ans
(a) \(P(F\cup S)=1-0.14(=0.86)\)
\(P(F\cap S)=0.93-0.86=0.07\)
Note: You can use Venn Diagram
(b) \(P(F|S) \left ( =\frac{P(F\cap S)}{P(S)} \right )=\frac{0.07}{0.62}=0.113\)
(c) \(F\) and \(S\) are not independent
EITHER
If independent \(P(F|S) = P(F), 0.113 \neq 0.31\)
OR
If independent \(P(F\cap S)=P(F)P(S), 0.07\neq 0.31\times 0.62(=0.1922)\)
(d) Let \(P(F) = x\)
\(P(S) = 2P(F) = 2x\)
\(P(F\cup S)=P(F)P(S)-P(F)P(S)\Leftrightarrow 0.86=x+2x-2x^{2}\Leftrightarrow 2x^{2}-3x+0.86=0\)
\(x = 0.386, x = 1.11\)
\(P(F) = 0.386\)
Question
The table below shows the subjects studied by \(210\) students at a college.
(a) A student from the college is selected at random.
Let \(A\) be the event the student studies Art.
Let \(B\) be the event the student is in Year \(2\).
(i) Find \(P(A)\).
(ii) Find the probability that the student is a Year \(2\) Art student.
(iii) Are the events \(A\) and \(B\) independent? Justify your answer.
(b) Given that a History student is selected at random, calculate the probability that the student is in Year \(1\).
(c) Two students are selected at random from the college. Calculate the probability that one student is in Year \(1\), and the other in Year \(2\).
Answer/Explanation
Ans
(a) (i) \(P(A)= \frac{80}{210}=\left ( \frac{8}{21}=0.381 \right )\)
(ii) \(P(year 2 art)=\frac{35}{210}=\left ( \frac{1}{6}=0.167 \right )\)
(iii) No (the events are not independent
EITHER \(P(A\cap B)=P(A)\times P(B)\) (to be independent)
\(P(B)=\frac{100}{210}\left ( =\frac{10}{21}=0.476 \right ) but \frac{1}{6}\neq \frac{8}{21}\times \frac{10}{21}\)
OR \(P(A)=P(A|B)\) (to be independent)
\(P(A|B)=\frac{35}{100}\) but \(\frac{8}{21}\neq \frac{35}{100}\)
OR \(P(A)=P(A|B)\) (to be independent)
\(P(B) = \frac{100}{210}\left ( =\frac{10}{21}=0.476 \right ), P(B|A)=\frac{35}{80}\) but \(\frac{35}{80}\neq \frac{100}{210}\)
(b) n(history) = \(85\)
\(P(year 1|history)=\frac{50}{85}=\left ( \frac{10}{17}=0.588 \right )\)
(c) \(\left ( \frac{110}{210}\times \frac{100}{209} \right ) + \left ( \frac{110}{210}\times \frac{100}{209} \right ) \left (= 2\times \frac{110}{210}\times \frac{100}{209} \right ) =\frac{200}{399} (=0.501)\)
Question
In a class of \(100\) boys, \(55\) boys play football and \(75\) boys play rugby. Each boy must play at least one sport from football and rugby.
(a) (i) Find the number of boys who play both sports.
(ii) Write down the number of boys who play only rugby.
(b) One boy is selected at random.
(i) Find the probability that he plays only one sport.
(ii) Given that the boy selected plays only one sport, find the probability that he plays rugby.
Let \(A\) be the event that a boy plays football and \(B\) be the event that a boy plays rugby.
(c) Explain why \(A\) and \(B\) are not mutually exclusive.
(d) Show that \(A\) and \(B\) are not independent.
Answer/Explanation
Ans
(a) (i) Venn diagram, \(30\)
(ii) \(45\)
(b) (i) \(\frac{70}{100}\left ( =\frac{7}{10} \right )\)
(ii) \(\frac{45}{70}\left ( =\frac{9}{14} \right )\)
(c) \(P(A \cap B) = 0.3 \neq 0\)
(d) \(P(A \cap B) \neq P(A) \times P(B)\), \(\frac{30}{100}\neq \frac{75}{100}\times \frac{55}{100}, \frac{30}{55} \neq \frac{75}{100}\)
OR \(P(B│A) \neq P(B)\), \(\frac{30}{55} \neq \frac{75}{100}\)
Question
An integer is chosen at random from the first one thousand positive integers.
Let \(A\) = {multiples of \(4\)} and \(B\) = {multiples of \(6\)} .
Then \(A \cap B\) = {multiples of \(12\)} (since the least common multiple of \(4\) and \(6\) is \(12\))
(a) Find the number of multiples
(i) of \(4\) (ii) of \(6\) (iii) of \(12\)
(b) Find the probability that the integer chosen is a multiple of \(4\).
(c) Find the probability that the integer chosen is a multiple of \(6\).
(d) Find the probability that the integer chosen is a multiple of both \(4\) and \(6\).
(e) Find the probability that the integer chosen is a multiple of \(4\) but not of \(6\).
(f) Find the probability that the integer chosen is a multiple of \(4\) or \(6\).
(g) Find the probability that the integer chosen is a multiple of \(4\) or \(6\) but not both.
Answer/Explanation
Ans
(a) (i) \(250\) (ii) \(166\) (since \(1000/6=166.67\)) (iii) \(83\)
(b) \(250/1000 = 1/4\)
(c) \(166/1000\)
(d) \(83/1000\)
(e) \(250-83 = 167\) so P=\(167/1000\).
(f) \(250 + 166 – 83 = 333\) so P=\(333/1000\)
(g) \(250 + 166 – 2×83 = 250\) so P=\(250/1000\).