Question

A fair coin is tossed 20 times. Let X be the number of heads thrown in the first 10 tosses, and let Y be the number of heads tossed in the last 10 tosses. Find the conditional probability that X = 6, given that X + Y = 10.

Answer 1

To find: P(X=6|X+Y=10)

Using the law of conditional probability,

P(X=6|X+Y=10) = P(X=6,Y=4)/P(X+Y=10)

Now,P(X=6,Y=4) = P(X=6)*P(Y=4)

{Since X and Y are independent}

= $\binom{10}{6}*(1/2)^{6}*(1/2)^{4} * \binom{10}{4}*(1/2)^{4}*(1/2)^{6}$ = 0.042057

Now, P(X+Y=10) = P(X=0,Y=10) + P(X=1,Y=9) + P(X=2,Y=8) +......... + P(X=9,Y=1) + P(X=10, Y=0)

= $\binom{20}{10}*(1/2)^{10}*(1/2)^{10}$

= 0.1762

Thus, required conditional probability = 0.042057/0.1762

= 0.2387