Question

Rohan throws an unbiased die. If he gets a $1,2,3,4,5$, he throws a coin and notes whether it is a head or a tail. But if he gets a $6$, he throws the coin four times and notes the number of heads. If he obtained exactly one head, then the probability that he threw a $6$ with the die is

(Assume the coin to be unbiased and each toss to be independent)

• A. $\frac{4}{9}$
• B. $\frac{1}{6}$
• C. $\frac{1}{11}$
• D. $\frac{5}{9}$

Answer 1

The correct option is C $\frac{1}{11}$

Let $E_1$ be the event of getting a $1,2,3,4,5$
and $E_2$ be the event of getting a $6$.
Then $P\left(E_1\right)=\frac{5}{6},P\left(E_2\right)=\frac{1}{6}$
Let $H$ be the event of getting head in a single toss.

By Bayes' theorem,
$P\left(E_2|1H\right)=\frac{P\left(1H|E_2\right)P\left(E_2\right)}{P\left(1H|E_1\right)P\left(E_1\right)+P\left(1H|E_2\right)P\left(E_2\right)}$

$=\frac{4{\left(\frac{1}{2}\right)}^4\times \frac{1}{6}}{\frac{1}{2}\times \frac{5}{6}+4{\left(\frac{1}{2}\right)}^4\times \frac{1}{6}}$

$=\frac{1}{11}$