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.
Given: n coins have head on both the sides
and (n + 1) coins are fair coins
Total coins = 2n + 1
Let events E1, E2 be the following:
E1 = Event that an unfair coin is selected
E2 = Event that a fair coin is selected
The Law of Total Probability:
In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that the toss result is a head
P(E|E1) is the probability of getting a head when unfair coin is tossed
P(E|E2) is the probability of getting a head when fair coin is tossed
So, 
From the law of total probability,
∴ P(E) = P(E1) × P(E|E1) + P(E2) × P(E|E2)
(Given)
⇒ 31 × 2(2n+1) = 42 × (3n + 1)
⇒ 124n + 62 = 126n + 42
⇒ 2n = 20
⇒ n = 10
Hence, the value of n is 10.