Question

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

Answer 1

Solution:-
The total number of ways one coin can be picked up from the bag of 2n + 1 coins = (2n + 1) c₁ =  2n +1
In the bag there are n coins with both sides have heads and therefore the number of fair coins = 2n + 1 - n
Fair coins = n + 1
Let A be the event that the unfair coins are tossed and that gives a head.
Number of ways one can pick an unfair coin is nc1 = n so, the probability of getting an unfair coin is n/(2n + 1).
Since the unfair coin tossed will always give a head
Therefore,
P (A) = 1*{n/(2n + 1)}
Now, let us consider in the same way the event B of getting a head from a fair coin.
In this case the probability of getting a fair coin is (n + 1)/(2n + 1)
However, in the fair coin the probability 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 probability of getting a head from either of the event is 
P (A ∪ B) = P (A) + P (B) = 31/42
Therefore, 
1/(2n + 1)*{n + (n + 1)/2} = 31/42
(3n + 1)/(4n + 2) = 31/42
126n + 42 = 124n + 62
126n - 124n = 62 - 42
2n = 20 
n = 10
Answer.