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 ride ?

Solution

Let A be the event that girl will get 1 or 2

$P (A) = \frac{2}{6} = \frac{1}{3}$

Let B be the event that girl will get 3,4, or 6

$P (B) = \frac{4}{6} = \frac{2}{3} P (\frac{T}{A}) = P r o b a b i l i t y o f e x a c t l y o n e t i l l g i v e n s h e g e t 1 o r 2 = \frac{3}{8} P (\frac{T}{B}) = P r o b a b i l i t y o f e x a c t l y o n e t i l l g i v e n s h e g e t 3, 4, 5 o r 6 = \frac{1}{2} P (\frac{B}{T}) = \frac{P (B) x P (\frac{T}{B})}{P (A) x P (\frac{T}{A}) + P (B) x P (\frac{T}{B})} = \frac{\frac{2}{3} x \frac{1}{2}}{\frac{1}{3} x \frac{3}{8} + \frac{2}{3} x \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{11}{8 x 3}} = \frac{8}{11}$

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