Question

if a fair coin is tossed five times, the probability of getting exactly two heads is

Explanation:

Step1: Use binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successful trials, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n = 5\) (coin - tosses), \(k = 2\) (number of heads), and \(p=\frac{1}{2}\) (probability of getting a head in a single coin - toss).
First, calculate \(C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5!}{2!3!}=\frac{5\times4\times3!}{2\times1\times3!}=10\).

Step2: Calculate the probability

Substitute \(n = 5\), \(k = 2\), \(p=\frac{1}{2}\) into the binomial formula:
\(P(X = 2)=C(5,2)\times(\frac{1}{2})^{2}\times(1-\frac{1}{2})^{5 - 2}\)
\(=10\times\frac{1}{4}\times(\frac{1}{2})^{3}\)
\(=10\times\frac{1}{4}\times\frac{1}{8}\)
\(=\frac{10}{32}=\frac{5}{16}\)

Answer:

\(\frac{5}{16}\)