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 $\frac{62}{42}$, then $n=$

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

Answer 1

The correct option is A $10$

$2n+1=$ total coins

there can be two cases

Case I - choosing biased coin

$P\left(B\right)=\frac{n}{2n+1}$

Case II = choosing unbiased coin

$P\left(UB\right)=\frac{n+1}{2n+1}$

$P\left(Head\right)=\frac{n}{2n+1}\times 1+\frac{n+1}{2n+1}\times \frac{1}{2}$

$=\frac{2n+(n+1)}{2(2n+1)}=\frac{3n+1}{2(2n+1)}$

$\Rightarrow \frac{3n+1}{2n+1}=\frac{62}{42}$

$\Rightarrow (3n+1)42=(2n+1)62$

$126n+42=124n+62$

$2n=62-42$

$2n=20$

$\Rightarrow n=10$