A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is m/n, gcd(m, n) = 1, then n² - m² is equal to :

Solution

Maths sir · 3 steps

Weight each coin by pick chance times head chance, then find the fair coins' share of the total.

1. Two possible coins. Let U be drawing an unbiased coin, D be drawing the double-headed coin, and H be getting a head. With 20 coins, P(U) = 19/20 and P(D) = 1/20.

   Coin	Chance of picking it	Chance of head	Picked and head
   Unbiased	19/20	1/2	19/40
   Double-headed	1/20	1	1/20 = 2/40

2. Apply Bayes. The total chance of a head is 19/40 + 2/40 = 21/40. The unbiased coins' share of that total is the probability we want.

   P(U | H) = (19/40)/(21/40) = 19/21

3. So. Since gcd(19, 21) = 1, m = 19 and n = 21. Then n² - m² = (21 - 19)(21 + 19) = 2 × 40 = 80, which is option (b).

The idea to keep. In Bayes questions, find the joint chance of each cause and the observed result, add them for the total, and take the share of the cause you are asked about.

Answer from the NTA final answer key. Solution by PrepWiser.