Solution: 2015 Winter Final - 16

Author: Michiel Smid

Question

Consider a bitstring of length 77, in which each bit is 0 with probability 1/3 (and, thus, 1 with probability 2/3), independently of the other bits. What is the probability that there are exactly 15 0s in this bitstring?

(a)

${77 \choose 15}(1/3)^{15}(2/3)^{62}$

(b)

${77 \choose 15}((1/3)^{62} + (2/3)^{15})$

(c)

${77 \choose 15}(1/3)^{62}(2/3)^{15}$

(d)

${77 \choose 15}((1/3)^{15} + (2/3)^{62})$

Solution

Let $X$ be the number of 0s in this bitstring.

$X \sim \text{Binomial}(77, 1/3)$

$ Pr(X=15) = \binom{77}{15} {\left( \frac{1}{3} \right)}^{15} {\left( \frac{2}{3} \right)}^{77-15} $