Question

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? 

Answer 1

Let E₁ : the event that 5 or 6 is shown on die and E₂ : the event that 1, 2, 3, or 4 is shown on die. 

Then, E₁ and E₂ are mutually exclusive and exhaustive events. 

and n(E₁) = 2, n(E₂) = 4 

Also, n(S) = 6

P(E₁) = 2/6 = 1/3 and P(E₂) = 4/6 = 2/3

Let E : The event that exactly one head show up,

∴ P(E/E₁) = P(exactly one head show up when coin is tossed thrice) = P{HTT, THT, TTH} = 3/8

P(E/E₂) = P(head shows up when coin is tossed once) = 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/E) 

By using Baye’s theorem, we obtain