Question

A box contains four coins. Three of the coins are fair, but one of them is biased, with P(11) = ? (where 11 is the event of flipping heads). You take a coin from the box and flip it. It comes up heads. What is the probability that you have flipped the biased coin?

Answer 1

H : Event of coming up with heads

E1 : Flipping a biased coin.

E2 : Flipping a fair coin

Probability that you have flipped the biased coin given you came up with heads: P(E1|H)

By bayes theorem,

There are four coins and one of the coin is biased, so probability of taking the baised coin from the box and flipping it = 1/4

P(E1) = 1/4

Similarly,

There are four coins and three of them are fair, so probability of taking the fair coin from the box and flipping it = 3/4

P(E2) = 3/4

Probability of getting heads given the flipped coin is baised = P(H|E1) = 3/4(Given)

Probability of getting heads given the fillped coin is fair = P(H|E2) = 1/2

Substituting the above values in the above formula

$\small P(E1|H) = \frac{P(E1)\times P(H|E1)}{P(E1)\times P(H|E1)+P(E2)\times P(H|E2)}=\frac{(1/4)\times (3/4)}{(1/4)\times (3/4)+(3/4)\times (1/2)}$

$\small =\frac{3/16}{(3/16)+(3/8)} = \frac{3/16}{9/16} = \frac{3}{9}=\frac{1}{3}$

Probability that you have flipped the biased coin given you came up with heads: P(E1|H) = $\small \frac{1}{3}$