Question

As a business strategy, $20 \%$ of the new internet service subscribers selected randomly receive a special promotion. If a group of $5$ such subscribers signs for the service, then the probability that at least two of them get the special promotion is

• A. $\frac{819}{3125}$
• B. $\frac{821}{3125}$
• C. $\frac{823}{3125}$
• D. $\frac{817}{3125}$

Answer 1

Answer: (B)

Here,
$n=5,\, P=20 \%=\frac{1}{5}$
$q=1-p=1-\frac{1}{5}=\frac{4}{5}$
Required probability
$P(r \geq 2)=P(r=2)+P(r=3)+P(r=4)+P(r=5)$
$P(r \geq 2) ={ }^{5} C_{2}\left(\frac{1}{5}\right)^{2}\left(\frac{4}{5}\right)^{3}$
$+{ }^{5} C_{3}\left(\frac{1}{5}\right)^{3}\left(\frac{4}{5}\right)^{2}+{ }^{5} C_{4}\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)+\left(\frac{1}{5}\right)^{5}$
$= \frac{1}{5^{5}}(10 \times 64+10 \times 16+5 \times 4+1)$
$= \frac{640+160+20+1}{3125}=\frac{821}{3125}$