When anunfaircoin is tossed twice, the probability of getting one tails and one heads is 1/2. What is the probability of getting two heads and two tails if the coin is tossed 4 times? Choose the correct choice
A) 2/3
B) 1/3
C) 1/4
D) 1/6
E) 1/9
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) = 1/2.   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) = 1/2 it follows that x*(1-x) = 1/4, then 4*(x*(1-x))^2 = 4*(1/6)^2 = 1/9.   Answer: 1/9.
<result>E</result>
When anunfaircoin is tossed twice, the probability of getting one tails and one heads is 1/2. What is the probability of getting two heads and two tails if the coin is tossed 4 times?