Question

Suppose a girl throws a die. If she gets a $5$ or $6$, she tosses a coin $3$ times and notes the number of heads. If she gets $1,2,3$ and $4$ 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 girl gets 5 or 6 on the roll. $P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$

Let $E_2$ be the event that the girl gets 1,2,3 or 4 on the roll. $P\left(E_2\right)=\frac{4}{6}=\frac{2}{3}$

Let A be the event that she obtained exactly one head.

If she tossed a coin 3 times and exactly 1 head showed up, then the total number of favourable outcomes = {(HTT), (THT), (TTH),} = 3.

$P\left(A|E_1\right)=\frac{3}{8}$ (Note: in 3 coin tosses, total sample space = 8)

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

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

We need to find the probability that she threw 1, 2, 3 or 4 with the die, given that she got exactly one head.

We can use Baye's theorem, according to which

$P\left(E_2|A\right)=\frac{P\left(E_2\right)\left(P\right(A|E_2)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}⟹=\frac{\frac{1}{2}\times \frac{2}{3}}{\frac{1}{2}\times \frac{2}{3}+\frac{3}{8}\times \frac{1}{3}}=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}=\frac{1}{3}\times \frac{24}{11}⟹P\left(E_2|A\right)=\frac{8}{11}$