The number of throws that the player make X follows binomial distribution.
The probability that the player makes x throws out of eleven throws is
[tex] P(X=x)=C(x,11)0.5^x(0.5)^{11-x}=C(x,11)0.5^{11} [/tex]
The probability that the player makes at most nine out of eleven free throws
is
[tex] \sum_{x=0}^9C(x,11)0.5^{11}=1-\sum_{x=10}^{11}C(x,11)0.5^{11}\\
=1-C(10,11)0.5^{11}-C(11,11)0.5^{11}=1-\frac{12}{2^{11}} \\
=1-\frac{3}{512} =\frac{509}{512} [/tex]
Correct choice is (B).
