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 $E_1$ be the event that the outcome on the die is $5$ or $6$ and $E_2$ be the event that the outcome on the die is $1,2,3$ or $4$.
$\therefore P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$ and $P\left(E_2\right)=\frac{4}{6}=\frac{2}{3}$
Let $A$ be the event of getting exactly one head.
$P\left(A|E_1\right)=$ Probability of getting exactly one head by tossing the coin three times if she gets $5$ or $6$ $=\frac{3}{8}$
$P\left(A|E_2\right)=$ Probability of getting exactly one head in a single throw of coin if she gets $1,2,3$ or $4$ $=\frac{1}{2}$
The probability that the girl threw $1,2,3$ or $4$ with the die, if she obtained exactly one head, is given by $P\left(E_2|A\right)$.
By using Bayes' theorem, we obtain
$P\left(E_2|A\right)=\frac{P\left(E_2\right)\cdot P\left(A|E_2\right)}{P\left(E_1\right)\cdot P\left(A|E_1\right)+P\left(E_2\right)\cdot P\left(A|E_2\right)}$
$=\frac{\frac{2}{3}\cdot \frac{1}{2}}{\frac{1}{3}\cdot \frac{3}{8}+\frac{2}{3}\cdot \frac{1}{2}}$
$=\frac{\frac{1}{3}}{\frac{1}{3}\left(\frac{3}{8}+1\right)}$
$=\frac{1}{\frac{11}{8}}$
$=\frac{8}{11}=0.72$