A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides
where as the rest of the coins are fair. A coin is picked up at random from the bag and is
tossed. If the probability that the toss results in a head is 31/42

, determine the value of n.

@Prashasti

The total number of ways one coin can be picked from a box of 2n + 1 coins is
(2n+1) C 1 = 2n+1

In this box there are n coins with both sides heads and therefore there are
2n+1 - n = n + 1 fair coins

Let A be the event that the unfair coin tossed gives a head.

No. of ways one can pick an unfair coin is --- n C 1 = n, thus the probablity of getting an unfair coin is
n/(2n+1)

Since the unfair coin tossed will always gives a head we have

P (A) = 1*(n/(2n+1))

Let's consider in the same way the event B of getting a heads from a fair coin

In this case the probablity of getting a fair coin is
(n+1)/(2n+1)

However in the fair coin the probablity of getting a head is 1/2, so

P (B) = (1/2)*(n+1)/(2n+1)

Now events A and B are mutually exclusive and the probablity of getting a head from either of the event is

P (A U B) = P (A) + P (B) = 31/42

So,

1/(2n+1)*(n + (n+1)/2) = 31/42

(3n + 1)/(4n + 2) = 31/42

which gives n = 10.

Hope the solution is clear.