The probability of a man hitting a target is (1)/(5). If the man...

Let $p$ be the probability of hitting the target and $q$ be the probability of missing the target.

$p = \dfrac{1}{5}$

$q = 1 - p = \dfrac{4}{5}$

Number of trials, $n = 7$

Let $X$ be the random variable representing the number of hits. The probability of hitting the target at least twice is given by:

$P(X \ge 2) = 1 - P(X \lt 2)$

$P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$

Using the binomial distribution formula $P(X = r) = ^{n}C_{r} p^r q^{n-r}$:

$P(X = 0) = ^{7}C_{0} \left(\dfrac{1}{5}\right)^0 \left(\dfrac{4}{5}\right)^7 = \left(\dfrac{4}{5}\right)^7$

$P(X = 1) = ^{7}C_{1} \left(\dfrac{1}{5}\right)^1 \left(\dfrac{4}{5}\right)^6 = \dfrac{7}{5} \left(\dfrac{4}{5}\right)^6$

Substituting these values:

$P(X \ge 2) = 1 - \left[ \left(\dfrac{4}{5}\right)^7 + \dfrac{7}{5} \left(\dfrac{4}{5}\right)^6 \right]<br><br>$P(X ≥ 2) = 1 - \left(\dfrac{4}{5}\right)^$6 \left[ \dfrac{4}{5} + \dfrac{7}{5} \right]$

$P(X \ge 2) = 1 - \left(\dfrac{11}{5}\right) \left(\dfrac{4}{5}\right)^6$

Answer: $1 - \left(\dfrac{11}{5}\right)\left(\dfrac{4}{5}\right)^6$