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

Both heads appear on n coins and head and a tail appear on (n+1)coins so
P(head) =$\frac{{}^n{C}_1}{2n+1_{C_1}}.1+\frac{n+1_{C_1}}{2n+1_{C_1}}.\frac{1}{2}=\frac{31}{42}$
$\Rightarrow \frac{n}{2n+1}+\frac{n+1}{2(2n+1)}=\frac{31}{42}\Rightarrow 2n+n+1=(2n+1)\left(\frac{31}{21}\right)\Rightarrow 63n+21=62n+31\Rightarrow n=10$.