Question

Suppose that you are given a bag containing n unbiased coins. You are told that n-1 of these coins are normal, with heads on one side and tails on the other, whereas one coin is a fake, with heads on both sides. Suppose you reach into the bag, pick out a coin at random, flip it, and get a head. What is the (conditional) probability that the coin you chose is the fake coin

Answer 1

The (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

Explanation:

Given

Total unbiased coin = n

Normal coins =n - 1

Fake = 1

The (conditional) probability that the coin you chose is the fake coin is represented by

P(Fake | Head)

And it's calculated as follows;

P(Fake | Head) = P(Fake, Head) ÷ P(Head) ----- (1)

Where P(Fake, Head) = P(Fake) * P(Head | Fake)

P(Fake) = 1/n --- because only one is fake

P(Head | Fake) = n/n because all coins (including the fake) have head

So, P(Fake, Head) = P(Fake) * P(Head | Fake) becomes

P(Fake, Head) = 1/n * n/n

P(Fake, Head) = 1/n

P(Head) is calculated by

P(Fake) * P(Head | Fake) + P(Normal) * P(Head | Normal)

P(Fake) * P(Head | Fake) = P(Fake, Head) = 1/n (as calculated above)

P(Normal) * P(Head | Normal) = ½ * (n - 1)/n ----- considering that the coin also has a tail with equal probability as that of the head.

Going back to (1)

P(Fake | Head) = P(Fake, Head) ÷ P(Head) becomes

P(Fake | Head) = (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ (1/n + (n - 1)/2n)

= (1/n) ÷ (2 + n - 1)/(2n)

= (1/n) ÷ (1 + n)/(2n)

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

= 2/(1 + n)

Hence, the (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)