The number of faulty resistors follows Binomial Distributions.

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

Every batch contains, on average $3\%$ of defective resistors. For a batch of $12$ resistors,

$($a$)$ what is the probability to be accepted if $2$ resistors are found?

$($b$)$ what is the probability to be accepted if $2$ or less defective are found?

$($c$)$ what is the probability to be accepted if more than $2$ defectives are found?

Round the answer in $4$ decimals places.