The probability of a man hitting a target is 1/5. If the man fires 7 times, then what is the probability that he hits the target at least twice?

Calculation:

Given,

Probability of hitting the target per shot, \(p = \frac{1}{5}\)

Number of shots fired, \(n = 7\)

The number of hits follows a binomial distribution:

\(X \sim \mathrm{Binomial}(n=7,\;p=\tfrac15)\)

Probability of zero hits:

\(P(X=0)=\left(\frac45\right)^{7}\)

Probability of exactly one hit:

\(P(X=1)=\binom71\!\left(\frac15\right)\!\left(\frac45\right)^{6} =\frac{7}{5}\left(\frac45\right)^{6}\)

Probability of at least two hits:

\(P(X\ge2)=1-\bigl[P(X=0)+P(X=1)\bigr] =1-\frac{11}{5}\left(\frac45\right)^{6} \)

∴ The probability of hitting the target at least twice is  \(1-\dfrac{11}{5}\left(\dfrac45\right)^{6} \).

Hence, the correct answer is Option 3.