Probability question help me please? A company produces monitors.30 % of the monitors do not work.What is the probability that in a box with 15 monitors.
a) 3 monitors dont work
b) at least 6 monitors dont work
c) less than 10 monitors dont work.
I am thinking about solving this with Bayes' theorem,but I have no Idea how to apply it here? Any help or hint? 
 A: I do not see a role for Bayes' Theorem.
We are intended to assume that with probability $0.30$, a randomly chosen monitor is defective, and that defectiveness of the various monitors in the box are independent events.
For (a), we are I think asked for the probability that exactly $3$ monitors are defective. The number of defective monitors in the box has binomial distribution, $n=15$, $p=0.30$. The probability that there are exactly $k$ defective is 
$$\binom{15}{k}p^k(1-p)^{15-k}.\tag{1}$$
From this you can quickly find the probability of $3$ defectives. 
For (b), we want the probability that the number of bads is $6$ or $7$ or $8$ or  $9$ or $\dots$. From (1) you can find the probability of $6$ bad, the probability of $7$ bad, and so on, and add  up.
It is easier to find the probability that the number of bads is $\lt 6$, by using (1) to find the probability of $0$ bad, $1$ bad, and so on up to $5$ bad. Add these up, and you get the probability of $5$ or fewer bad. From this you should be able to write down the probability of $6$ or more bad.
The last problem does not involve any new idea.
A: Some hints:


*

*What is the probability of a monitor failure?

*Make the assumption that a monitor's failure is independent of any other monitor's failure.

*If you have two monitors then the probability that one fails and the other does not is ....
