Question

A bag contains $(2 n+1)$ 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 $31 / 42$, then $n$ is equal to

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

The Correct Option is A

Total number of coins $=2 n +1$
Consider the following events:
$E_{1}=$ Getting a coin having head on both sides from the bag.
$E_{2}=$ Getting a fair coin from the bag
$A =$ Toss results in a head
Given: $P(A)=\frac{31}{42}, P\left(E_{1}\right)=\frac{n}{2 n+1}$
and $P\left(E_{2}\right)=\frac{n+1}{2 n+1}$
Then, $P(A)=P\left(E_{1}\right) P\left(A E_{1}\right)+P\left(E_{2}\right) P\left(A / E_{2}\right)$
$\Rightarrow \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2}$
$\frac{31}{42}=\frac{n}{2 n+1}+\frac{n+1}{2(2 n+1)}$
$\Rightarrow \frac{31}{42}=\frac{3 n+1}{2(2 n+1)}$
$\Rightarrow \frac{31}{21}=\frac{3 n+1}{2 n+1}$
$n=10$