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? [CBSE 2015]

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,

$\mathrm{P}\left(E_1\right)=\frac{2}{6}=\frac{1}{3}\mathrm{and}\mathrm{P}\left(E_2\right)=\frac{4}{6}=\frac{2}{3}$

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

$$\begin{gathered} \mathrm{P}\left(E_2|A\right)=\frac{\mathrm{P}\left(E_2\right)\times \mathrm{P}\left(A|E_2\right)}{\mathrm{P}\left(E_1\right)\times \mathrm{P}\left(A|E_1\right)+\mathrm{P}\left(E_2\right)\times \mathrm{P}\left(A|E_2\right)} \\ =\frac{\left({\displaystyle \frac{2}{3}}\times {\displaystyle \frac{1}{2}}\right)}{\left({\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{8}}+{\displaystyle \frac{2}{3}}\times {\displaystyle \frac{1}{2}}\right)} \\ =\frac{\left({\displaystyle \frac{2}{6}}\right)}{\left({\displaystyle \frac{1}{8}+\frac{1}{3}}\right)} \\ =\frac{\left({\displaystyle \frac{2}{6}}\right)}{\left({\displaystyle \frac{11}{24}}\right)} \\ =\frac{24\times 2}{11\times 6} \\ =\frac{8}{11} \end{gathered}$$

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