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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?

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Why do you think Bayes theorem will be useful? – response Jun 1 '13 at 18:21
Intuition........ – hgdhg Jun 1 '13 at 18:22
Intuition is a good guide but it needs to be backed by checking the math. – response Jun 1 '13 at 18:24

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.

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Some hints:

  1. What is the probability of a monitor failure?

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

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

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