Bunuel wrote:

When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3. What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A. 2/3

B. 1/3

C. 1/4

D. 1/6

E. 1/9

OFFICIAL SOLUTION:Let the probability of heads be x and the probability of tails be 1-x.

So, we are given that P(one tails and one heads) = 2*x*(1-x) = 1/3.

P(two heads and two tails) = 4!/(2!2!)*x^2*(1-x)^2 = 6*(x*(1-x))^2. We are multiplying by 4!/(2!2!) because HHTT scenario can occur in several ways: HHTT, HTHT, THHT, ... (permutation of 4 letters HHTT where 2 T's and 2 H's are the same).

Since from 2*x*(1-x) = 1/3 it follows that x*(1-x) = 1/6, then 6*(x*(1-x))^2 = 6*(1/6)^2 = 1/6.

Answer: D.

whats difference between fair and unfair coin??...