WJEC Maths for A2 – Applied

1 Probability 14 4 Jack is taking part in a quiz programme. For each question in the quiz, four alternative answers are given, only one of which is correct. Jack has probability 0.6 of knowing the correct answer to a question, and when he does not know the correct answer, he chooses one of the four answers at random. (a) Calculate the probability that Jack gives the correct answer to a question. (b) Given that Jack gave the correct answer to a question, ϐind the probability that he knew the correct answer. Answer 4 (a) Let K = event knowing the correct answer A = event giving the correct answer K K Ԣ P( K ) = 0.6 P( K Ԣ ) = 0.4 P( A | K ) = 1 P( A | K Ԣ ) = 0.25 P( A Ԣ | K Ԣ ) = 0.75 A A A Ԣ P(giving correct answer) = (0.6 × 1) + (0.4 × 0.25) = 0.7 (b) P(knew correct answer|gave the correct answer) = Pȍknew the correct answerȎ P(gave correct answer) = 0.6 0.7 = 0.857(correct to 3 d.p.) Determining whether two events are independent or not Suppose we have two events A and B and we want to ϐind out if they are independent events (i.e. the occurrence of one event has no inϐluence on the other event occurring) or dependent events, we can use the method shown here. If the two events are independent, then the multiplication law for independent events is true, so: P( A ∩ B ) = P( A ) × P( B ) The generalised addition law, shown below, can be used for dependent or independent events. P( A ∩ B ) = P( A ) + P( B ) − P( A ∪ B ) Hence if you found the value of P( A ∩ B ) using the generalised addition law and the value was not the same as that found using the multiplication law, then the events A and B are dependent. It is always advisable to draw a tree diagram unless the number of branches makes it too difϐicult to draw. Notice that once Jack knows the correct answer, the probability then of him giving the correct answer is 1. There are two paths to consider on the tree diagram.