A manufacturer of household devices reports that on average $10$% of its products are defective. In a random sample of $5$, what is the probability that:
$P(D)=0.1$ $P(D^c)=0.9$
(a) Exactly $3$ devices are defective
C$5\choose 3$ = $\frac{5!}{3!}$ but I don't know where to implement the appropriate probabilities
(b) None are defective
C$5\choose 0$
I don't know the other two at all
(c) Less than $2$ devices are defective
(d) More than $3$ devices are defective
The notation should be either $\mathrm C^5_3$ or $\binom 5 3$, not a mixture. It is also equal to $\tfrac {5!}{3!~2!}$.
The binomial coefficien, $\mathrm C^5_3$, counts the arrangements of 3 defectives among a sample of 5 (alternatively: ways to select places for the defectives in a line up of the samples). You also need to measure the probability for a particular arrangement of three defectives and two nondefectives to occur. Since each such arrangement is equally probable, multiplying that by the coefficient gives the probability that some arrangement of three defectives and two nondefectives will occur.
$${\mathrm C}^5_3~\mathsf P(D)^3~\mathsf P(D^\complement)^2$$