Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solution
Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.
`:. P(E_1) = 2/6 = 1/3 and P(E_2) = 4/6 = 2/3`
Let A be the event of getting exactly one head.
P (A|E1) = Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6 = 3/8
P (A|E2) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4 = 1/2
The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E2|A).
By using Bayes’ theorem, we obtain
