Question

Consider 5 independent Bernoulli trials each with probabilities of success $\frac{1}{2}$, then the probability that number of success are more than the number of failure is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{2}{5}$

Answer 1

Answer: (C)

$\frac{1}{2}$

Since success and failure have the same chance of occurence. Hence corresponding to every case of $mS+nF$ there is a case of $mF+nS$ where $S=$ success, $F=$ Failure.
Hence $P(m>n)=\frac{1}{2}$
(Note $m=n$ is not possible)