it is reported that 72% of the working women use computers at work. Choose 5 women at random and find the probability that at least 1 doesn't use a computer at work

Let the random variable $X$ be the number of women who do not use computer at work. $X$ follows a Binomial distribution with parameters, $n=5$ and $p=1-0.72=0.28$. That is, $X\sim Bin(n=5,p=0.28)$ and it is given as,

$p(X=x)=\binom{5}{x}0.28^x(1-0.28)^{5-x},\space x=0,1,2,3,4,5$

Probability that at least 1woman does not use a computer at work is given as,

$p(x\geq1)=\displaystyle\sum^5_1p(X=x)$. To make it simpler, this probability is equivalent to,

$p(x\geq1)=1-p(X=0)$

So,

$p(X=0)=\binom{5}{0}0.28^0(0.72)^5=0.72^5=0.1934918(4dp)$.

Thus,

$p(x\geq1)=1-p(X=0)=1-0.1934918= 0.8065082$

Therefore, the probability that at least 1 woman does not use a computer at work is  0.8065082.