Question

A bag contains 2n+1 coins , It is known that n of these coins have ahead 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=

A) 10
B) 11
C) 12
D) 13

Answer 1

Given that there are two types of coins in the bag. So, in order to find the probability that the toss results in a head , we need to find the probabilities of the two cases that is the coin is from that n coins or the coins is from that n+1 coins and then find the probability that it will be heads and add both the probability cases.

Complete step by step answer:
Given that the n out of 2n+1 coins have heads on both sides, which implies the probability that the toss will be heads from these coins will be 1.
And as the other n+1 coins are fair , that implies the probability that the toss will be heads from these coins will be 1/2.
Now the probability the coin drawn will be from one of those n coins will be n/2n+1.
Similarly, the probability the coin drawn will be from one of those n coins will be n+1/2n+1.
Given that the probability that the toss results in a head is 31/42
=> (Probability that the coin is drawn from n coins× probability that the toss is head in these n coins) + (Probability that the coin is drawn from n+1 coins × probability that the toss is head in these n+1 coins) = 31/42
=> $(\dfrac{n}{{2n + 1}} \times 1) + (\dfrac{{n + 1}}{{2n + 1}} \times \dfrac{1}{2})$ = $\dfrac{{31}}{{42}}$
=> $\dfrac{{3n + 1}}{{2n + 1}} = \dfrac{{31}}{{21}}$
=> $63n + 21 = 62n + 31$
=> n=10
Therefore the correct option is A.

Note:
Read the question correctly and find whether they have asked the value of n or the total balls in the bag , because the total balls in the bag will be 21. And you can also recheck the question with the answer you get by backtracking method if you are not sure with your answer.