Question
Andre will play in the semi-final of a tennis tournament.
If Andre wins the semi-final he will progress to the final. If Andre loses the semi-final,
he will not progress to the final.
If Andre wins the final, he will be the champion.
The probability that Andre will win the semi-final is p . If Andre wins the semi-final, then the
probability he will be the champion is 0.6.
Complete the values in the tree diagram. [1]
- b. The probability that Andre will not be the champion is 0.58.
- c. Find the value of p . [2]
- d. Given that Andre did not become the champion, find the probability that he lost in the semi-final. [3]
Answer/Explanation
Ans:
(a) 
(b)
p × 0.4+ (1-9)=0.58
OR
P × 0.6 = 1- 0.58
P= 0.7
(c)
\(\frac{0.3}{0.58}(\frac{1-0.7}{0.58})\)
OR
\(\frac{0.3}{0.3+0.7\times 0.4}\)
\(\frac{15}{29}\)(0.517,0.517241…,51.7%)
Question
The diagram below shows the probabilities for events A and B , with \({\rm{P}}(A’) = p\) .
Write down the value of p .
Find \({\rm{P}}(B)\) .
Find \({\rm{P}}(A’|B)\) .
Answer/Explanation
Markscheme
\(p = \frac{4}{5}\) A1 N1
[1 mark]
multiplying along the branches (M1)
e.g. \(\frac{1}{5} \times \frac{1}{4}\) , \(\frac{{12}}{{40}}\)
adding products of probabilities of two mutually exclusive paths (M1)
e.g. \(\frac{1}{5} \times \frac{1}{4} + \frac{4}{5} \times \frac{3}{8}\) , \(\frac{1}{{20}} + \frac{{12}}{{40}}\)
\({\rm{P}}(B) = \frac{{14}}{{40}}\) \(\left( { = \frac{7}{{20}}} \right)\) A1 N2
[3 marks]
appropriate approach which must include \({A’}\) (may be seen on diagram) (M1)
e.g. \(\frac{{{\rm{P}}(A’ \cap B)}}{{{\rm{P}}(B)}}\) (do not accept \(\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}\) )
\({\rm{P}}(A’|B) = \frac{{\frac{4}{5} \times \frac{3}{8}}}{{\frac{7}{{20}}}}\) (A1)
\({\rm{P}}(A’|B) = \frac{{12}}{{14}}\) \(\left( { = \frac{6}{7}} \right)\) A1 N2
[3 marks]
Question
Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
(i) Copy and complete the following tree diagram.
(ii) Find the probability that two white balls are chosen.
Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is \(\frac{2}{7}\) .
A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.
Find the probability that the two balls are white.
Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probability that they are both white is \(\frac{2}{7}\) .
A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bag B.
Given that both balls are white, find the probability that they were chosen from bag A.
Answer/Explanation
Markscheme
(i)
\(\frac{4}{6},\frac{3}{6}{\rm{and}}\frac{3}{6}\left( {\frac{2}{3},\frac{1}{2}{\rm{and}}\frac{1}{2}} \right)\) A1A1A1 N3
(ii) multiplying along the correct branches (may be seen on diagram) (A1)
e.g. \(\frac{3}{7} \times \frac{2}{6}\)
\(\frac{6}{{42}}\left( { = \frac{1}{7}} \right)\) A1 N2
[5 marks]
\({\rm{P(bag A) = }}\frac{2}{6}\left( { = \frac{1}{3}} \right)\) , \({\rm{P(bag B) = }}\frac{4}{6}\left( { = \frac{2}{3}} \right)\) (seen anywhere) (A1)(A1)
appropriate approach (M1)
e.g. \({\rm{P(}}WW \cap A) + {\rm{P}}(WW \cap B)\)
correct calculation A1
e.g. \(\frac{1}{3} \times \frac{1}{7} + \frac{2}{3} \times \frac{2}{7}\) , \(\frac{2}{{42}} + \frac{8}{{42}}\)
\({\rm{P}}(2W) = \frac{{60}}{{252}}\left( { = \frac{5}{{21}}} \right)\) A1 N3
[5 marks]
recognizing conditional probability (M1)
e.g. \(\frac{{{\rm{P}}(A \cap B)}}{{{\rm{P}}(B)}}\) , \({\rm{P}}(A|WW) = \frac{{{\rm{P}}(WW \cap A)}}{{{\rm{P}}(WW)}}\)
correct numerator (A1)
e.g. \({\rm{P}}(A \cap WW) = \frac{6}{{42}} \times \frac{2}{6},\frac{1}{{21}}\)
correct denominator (A1)
e.g. \(\frac{6}{{252}},\frac{5}{{21}}\)
probability \(\frac{{84}}{{420}}\left( { = \frac{1}{5}} \right)\) A1 N3
[4 marks]
Question
In a class of 21 students, 12 own a laptop, 10 own a tablet, and 3 own neither.
The following Venn diagram shows the events “own a laptop” and “own a tablet”.
The values \(p\), \(q\), \(r\) and \(s\) represent numbers of students.
A student is selected at random from the class.
Two students are randomly selected from the class. Let \(L\) be the event a “student owns a laptop”.
(i) Write down the value of \(p\).
(ii) Find the value of \(q\).
(iii) Write down the value of \(r\) and of \(s\).
(i) Write down the probability that this student owns a laptop.
(ii) Find the probability that this student owns a laptop or a tablet but not both.
(i) Copy and complete the following tree diagram. (Do not write on this page.)
(ii) Write down the probability that the second student owns a laptop given that the first owns a laptop.
Answer/Explanation
Markscheme
(i) \(p = 3\) A1 N1
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)\((12 + 10 + 3) – 21,{\text{ }}22 – 18\)
\(q = 4\) A1 N2
(iii) \(r = 8,{\text{ }}s = 6\) A1A1 N2
(i) \(\frac{{12}}{{21}}{\text{ }}\left( { = \frac{4}{7}} \right)\) A2 N2
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)\(8 + 6,{\text{ }}r + s\)
\(\frac{{14}}{{21}}{\text{ }}\left( { = \frac{2}{3}} \right)\) A1 N2
(i) 
A1A1A1 N3
(ii) \(\frac{{11}}{{20}}\) A1 N1
[4 marks]