Question

A bag contains $(2n+1)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $(n+1)$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is A $10$

$E_1$: biased coin is chosen,
$E_2$: fair coin is choice.
and A: toss results in a head.
$P\left(E_1\right)=n/(2n+1),P\left(E_2\right)=(n+1)/(2n+1)$
$P\left(A|E_1\right)=1$ and $P\left(A|E_2\right)=1/2$
By the total probability rule
$\frac{31}{42}=\frac{n}{2n+1}\left(1\right)+\frac{(n+1)}{2(2n+1)}\Rightarrow \frac{3n+1}{2n+1}=\frac{31}{21}\Rightarrow n=10$