Question

A bag contains $(2n+1)$ coins, It is known that n of these coin have a head on both sides, whereas the remaining $(n+1)$ coins are fair. A coin is selected at random from the bag and tossed once. If the probability 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 $10$

$\frac{31}{42}=\left(\frac{n}{2n+1}\right)\cdot 1+\frac{(n+1)}{(2n+1)}\cdot \frac{1}{2}$
$\Rightarrow \left(31\right)\cdot 2(2n+1)=42(2n+n+1)$
$\Rightarrow 31(2n+1)=21(3n+1)$
$\Rightarrow 62n+31=63n+21$
$\Rightarrow n=10$.