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 selected at random from the bag and tossed once. If the probability that the toss results in a head is $\frac{31}{42},$ then $n$ is equal to

• A. $12$
• B. $10$
• C. $7$
• D. $8$

Answer 1

The correct option is B $10$

Probability of choosing biased coin $=\frac{n}{2n+1}$
Probability of choosing unbiased coin $=\frac{n+1}{2n+1}$
$P\left(\text{Head}\right)=\left(\frac{n}{2n+1}\right)\cdot 1+\left(\frac{n+1}{2n+1}\right)\cdot \frac{1}{2}$
$\Rightarrow \frac{31}{42}=\left(\frac{n}{2n+1}\right)\cdot 1+\left(\frac{n+1}{2n+1}\right)\cdot \frac{1}{2}$
$\Rightarrow \left(31\right)\cdot 2(2n+1)=42(2n+n+1)$
$\Rightarrow 31(2n+1)=21(3n+1)$
$\Rightarrow 62n+31=63n+21$
$\Rightarrow n=10$