Question

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

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

Answer 1

Answer: (B)

$1-{\left(\frac{3}{4}\right)}^7$

Let the probability of hitting the target $n$ times out of 7 be $P(N=n)$.
Probability of not hitting the target in one trial $=1-\frac{1}{4}=\frac{3}{4}$
$\therefore$ Probability of hitting the target $0$ times out of $7$ $=P(N=0)={\left(\frac{3}{4}\right)}^7$
Now, hitting the target $0$ times out of $7$ and hitting it at least once are mutually exclusive events.
$\therefore P(N=0)+P(N\ge 1)=1$
$\Rightarrow$ P (hitting target at least once)$=P(N\ge 1)=1-{\left(\frac{3}{4}\right)}^7$
Hence, option 'B' is correct.