Question

The probability of a man hitting the target is $\frac{1}{4}$. If he fires 7 times, the probability of hitting the target exactly six times is

• A. ${}^7{C}_5{\left(\frac{1}{4}\right)}^6\left(\frac{3}{4}\right)$
• B. ${}^7{C}_6{\left(\frac{3}{4}\right)}^6\left(\frac{1}{4}\right)$
• C. ${}^7{C}_6{\left(\frac{1}{4}\right)}^6\left(\frac{3}{4}\right)$
• D. ${}^7{C}_6{\left(\frac{1}{2}\right)}^6\left(\frac{1}{2}\right)$

Answer 1

The correct option is C ${}^7{C}_6{\left(\frac{1}{4}\right)}^6\left(\frac{3}{4}\right)$

The probability of a man hitting the target is $\frac{1}{4}$
For fires 7 times, $n=7$
The probability of hitting the target exactly six times is = $P(X=k){=}^n{C}_k{p}^k(1-p{)}^{n-k}$

$P(X=6)={}^7{C}_6{\left(\frac{1}{4}\right)}^6{\left(1-\frac{1}{4}\right)}^{7-6}$

$={}^7{C}_6{\left(\frac{1}{4}\right)}^6\left(\frac{3}{4}\right)$