ANSWER ALL QUESTIONS.

Every batch contains $15\%$ of defective screws on average. For a set of $10$ screws, using Binomial Distributions,

$P(r)=\frac{n!}{(n-r)!r!}{p}^r(1-p{)}^{n-r}$

determine the probability that

$($a$)$ exactly $4$ defectives.

$($b$)$ at most $4$ defectives.

$($c$)$ at least $4$ defectives.

Round the answer to $4$ decimals places.