#### Question

Suppose a girl throws a die. If she gets 1 or 2 she tosses a coin three times and notes the number of tails. If she gets 3,4,5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3,4,5 or 6 with the die ?

#### Solution

Let E_{1} be the event that the outcome on the die is 1 or 2 and E_{2} be the event that the outcome on the die is 3, 4, 5 or 6. Then,

`P(E_1) = 2/6 = 1/3 and P(E_2) = 4/6 = 2/3`

Let *A* be the event of getting exactly one 'tail'.

P(A|E_{1}) = Probability of getting exactly one tail by tossing the coin three times if she gets 1 or 2 = `3/8`

P(A|E_{2}) = Probability of getting exactly one tail in a single throw of a coin if she gets 3, 4, 5 or 6 = `1/2`

As, the probability that the girl threw 3, 4, 5 or 6 with the die, if she obtained exactly one tail, is given by P(E_{2}|A).

So, by using Baye's theorem, we get

`P(E_2|A) = (P(E_2)xxP(A|E_2))/(P(E_1)xxP(A|E_1)+P(E_2) xx P(A|E_2))`

`= (2/3 xx 1/2)/((1/3xx3/8+2/3xx1/2))`

`= (2/6)/((1/8","+","1/3))`

`= ((2/6))/((11/24))`

`= (24xx2)/(11xx6)`

`= 8/11`

So, the probability that she threw 3, 4, 5 or 6 with the die if she obtained exactly one tail is `8/11`