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June 2010 p61 q7
2636
In a television quiz show Peter answers questions one after another, stopping as soon as a question is answered wrongly.
The probability that Peter gives the correct answer himself to any question is 0.7.
The probability that Peter gives a wrong answer himself to any question is 0.1.
The probability that Peter decides to ask for help for any question is 0.2.
On the first occasion that Peter decides to ask for help he asks the audience. The probability that the audience gives the correct answer to any question is 0.95. This information is shown in the tree diagram below.
(i) Show that the probability that the first question is answered correctly is 0.89.
On the second occasion that Peter decides to ask for help he phones a friend. The probability that his friend gives the correct answer to any question is 0.65.
(ii) Find the probability that the first two questions are both answered correctly.
(iii) Given that the first two questions were both answered correctly, find the probability that Peter asked the audience.
Solution
(i) The probability that the first question is answered correctly is calculated as follows:
\(P(\text{1st correct}) = P(\text{Peter correct}) + P(\text{Peter asks for help and audience correct})\)
\(= 0.7 + 0.2 \times 0.95 = 0.89\)
(ii) To find the probability that the first two questions are both answered correctly, consider the following scenarios:
Peter answers both questions correctly: \(P(CC) = 0.7 \times 0.7 = 0.49\)
Peter answers the first correctly, asks for help, and the audience answers correctly: \(P(CHA) = 0.7 \times 0.2 \times 0.95 = 0.133\)
Peter asks for help on the first, audience answers correctly, and Peter answers the second correctly: \(P(HAC) = 0.2 \times 0.95 \times 0.7 = 0.133\)
Peter asks for help on the first, audience answers correctly, Peter asks for help on the second, and friend answers correctly: \(P(HAHP) = 0.2 \times 0.95 \times 0.2 \times 0.65 = 0.0247\)
