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?

Answer 1

Let $E_1$ be the event that the girl gets $1$ or $2$

Thus, $P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$

Let $E_2$ be the event that the girl gets $3,4,5$ or $6$

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

Let $A$ be the event that she obtained exactly one tail.

If she tossed a coin $3$ times and exactly $1$ tail showed up, then the total number of favorable outcomes $=\left\{\right(THH),(HTH),(HHT\left)\right\}=3$.

Thus $P\left(A|E_1\right)=\frac{3}{8}$

If she tossed the coin only once and exactly $1$ tail showed up, the total number of favorable outcomes $=1$.

Thus $P\left(A|E_2\right)=\frac{1}{2}$

The probability that she threw $3,4,5$ or $6$ with the die, given that she got exactly one tail can be found as follows:

Or in other words, we need to find $P\left(E_2|A\right)$

We can use the Baye's theorem, according to which $P\left(E_2|A\right)=\frac{P\left(E_2\right)\left(P\right(A|E_2)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$

$P\left(E_2|A\right)=\frac{\frac{1}{2}\cdot \frac{2}{3}}{\frac{1}{2}\cdot \frac{2}{3}+\frac{3}{8}\cdot \frac{1}{3}}=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}=\frac{8}{11}$