Products from a certain machine are too large 15% of the time. What is the...

${\displaystyle C_5^{20}{\left(.15\right)}^5{\left(.85\right)}^{15}\approx .103}$
Explanation:
The probability that 5 parts are too large is ${\displaystyle .15\times .15\times .15\times .15\times .15={\left(.15\right)}^5}$
and since 5 parts are too large, the other 15 are of acceptable size with probability
${\displaystyle {(1-.15)}^{15}.}$ Since any 5 parts can be too large in no particular order, there are ${\displaystyle C_5^{20}}$
ways to do this, so the probability that exactly 5 parts are too large is
${\displaystyle C_5^{20}{\left(.15\right)}^5{\left(.85\right)}^{15}\approx .103.}$

Given, product from a certain machine too large
is 15% of time i.e p=0.15
${\displaystyle \therefore }$ q=1-p=1-0.15=0.85
sample size (n)=20
Binomial distribution: ${\displaystyle P(X=\gamma )=n_{C_{\gamma }}{p}^{\gamma }{q}^{n-\gamma }}$
${\displaystyle \therefore p(X=5)={20}_{C_5}{\left(0.15\right)}^5{\left(0.85\right)}^{15}=0.1028}$
Answer = 0.1028