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', then what is the probability that she threw 3, 4, 5 or 6 with the die?       

Answer 1

Let E₁ be the event that the outcome on the die is 1 or 2 and E₂ be the event that the outcome on the die is 3, 4, 5 or 6. Then,

\[P\left( E_1 \right) = \frac{2}{6} = \frac{1}{3} \text{ and }  P\left( E_2 \right) = \frac{4}{6} = \frac{2}{3}\]

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

P(A|E₁) = Probability of getting exactly one tail by tossing the coin three times if she gets 1 or 2 = \[\frac{3}{8}\]

P(A|E₂) = Probability of getting exactly one tail in a single throw of a coin if she gets 3, 4, 5 or 5 = \[\frac{1}{2}\]

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(E₂|A).

So, by using Baye's theorem, we get

\[P\left( E_2 |A \right) = \frac{P\left( E_2 \right) \times P\left( A| E_2 \right)}{P\left( E_1 \right) \times P\left( A| E_1 \right) + P\left( E_2 \right) \times P\left( A| E_2 \right)}\]

\[ = \frac{\left( \frac{2}{3} \times \frac{1}{2} \right)}{\left( \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2} \right)}\]

\[ = \frac{\left( \frac{2}{6} \right)}{\left( \frac{1}{8} + \frac{1}{3} \right)}\]

\[ = \frac{\left( \frac{2}{6} \right)}{\left( \frac{11}{24} \right)}\]

\[ = \frac{24 \times 2}{11 \times 6}\]

\[ = \frac{8}{11}\]

So, the probability that she threw 3, 4, 5 or 6 with the die if she obtained exactly one tail is \[\frac{8}{11}\]