The probability of a man hitting a target is 0.25. If he shoots 7 times, then what is the probability of his hitting atleast twice?

Here, $n = 7p = 0.25 = \frac{1}{4},q = 1 - \frac{1}{4} = \frac{3}{4}r \ge 2$

where, $P(X){ = ^n}{C_r}{(p)^r}{(q)^{n - r}}$

In this case for easy approach we shall first find out the probability of his hitting atmost once (i.e., $r = 0,1$)

and then subtract this probability from 1 to get the desired probability.

$\therefore$ $P(X = r) = 1 - [P(r = 0) + P(r = 1)]$

$= 1 - \left[ {^7{C_0}{{\left( {\frac{1}{4}} \right)}^0}{{\left( {\frac{3}{4}} \right)}^{7 - 0}}{ + ^7}{C_1}{{\left( {\frac{1}{4}} \right)}^1}{{\left( {\frac{3}{4}} \right)}^{7 - 1}}} \right]$

$= 1 - \left[ {\frac{{7!}}{{0!7!}}{{\left( {\frac{3}{4}} \right)}^7} + \frac{{7!}}{{1!6!}}\left( {\frac{1}{4}} \right){{\left( {\frac{3}{4}} \right)}^6}} \right]$

$= 1 - \left[ {{{\left( {\frac{3}{4}} \right)}^6}\left( {\frac{3}{4} \cdot 1 + \frac{1}{4} \cdot 7} \right)} \right]$

$= 1 - \left[ {\frac{{{3^6}}}{{{4^6}}}\left( {\frac{{10}}{4}} \right)} \right] = 1 - \left[ {\frac{{{3^6} \times 10}}{{{4^7}}}} \right] = 1 - \left[ {\frac{{27 \cdot 27 \cdot 10}}{{64 \cdot 256}}} \right]$

$= 1 - \left[ {\frac{{7290}}{{16384}}} \right] = 1 - \frac{{3645}}{{8192}} = \frac{{4547}}{{8192}}$

Using Binomial distribution $P(X = r)$

to get the answer.

Here,
$P(X = r) = 1 - [P(r = 0) + P(r = 1)]$