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??...