Question

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides whereas remaining 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 value of n be

• 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 tail appear on (n + 1) coins,
therefore P(getting head) $=\frac{n{C}_1}{(2n+1)}{C}_1.1+\frac{(n+1)}{(2n+1)}.\frac{1}{2}=\frac{31}{42}\Rightarrow \frac{n}{2n+1}+\frac{n+1}{2(2n+1)}=\frac{31}{42}\Rightarrow 63n+21=62n+31\Rightarrow n=10.$