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
$P(r\ge 2)=P(r=2)+P(r=3)+P(r=4)+P(r=5)$
$P(r\ge 2)={}^5{C}_2{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^3$
$+{}^5{C}_3{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^2+{}^5{C}_4{\left(\frac{1}{5}\right)}^4\left(\frac{4}{5}\right)+{\left(\frac{1}{5}\right)}^5$
$=\frac{1}{5^5}(10\times 64+10\times 16+5\times 4+1)$
$=\frac{640+160+20+1}{3125}=\frac{821}{3125}$