The weights of a population of mice fed a certain diet follow a normal distribution with mean μ=100 grams and standard deviation σ=20 grams. A random sample of 8 such mice is taken.

(a) Find the probability that exactly 4 of the mice weigh between 80 and 100 grams, 2 of the mice weigh more than 100 grams, and 2 mice have weights less than 80 grams.

How I would do it, is say 4/8 are in 34% range, 2/8 are in 50% range, and 2/8 are in 16% range so I would average those numbers and get 33.5% but I'm not sure if that is right. Could someone let me know if I did it right.


  • $\begingroup$ Have you attempted all the parts? For each one, please tell us what you tried and where you got stuck. In the future, please only ask one question at a time. (One advantage of asking one at a time is that maybe after question $n$ gets answered, you will be able to figure out question $n+1$ on your own.) $\endgroup$ – Trevor Wilson Oct 16 '13 at 21:31

You have the right probabilities for each weight range (to two decimal places, anyway,) but you are not combining them in the right way. It doesn't make sense to average them (e.g. if I have a $99.999\%$ chance of not winning the lottery this week and a $0.001\%$ chance of winning the lottery next week, that doesn't mean that I have a $50\%$ chance of winning the lottery on exactly one of the two weeks.)

To get the probability that the event $A$ happens and an independent event $B$ also happens, you multiply the probabilities: $P(A \mathbin{\&} B) = P(A)P(B)$.

So the probability that, out of your eight mice, the first four mice are in the $80$–$100$ gram range, the next two mice have mass $>100$ grams, and the last two mice have mass $<80$ grams, is given by

\begin{align*}\tag{$\ast$}(.34)^4 (.50)^2 (.16)^2.\end{align*}

However, this is not yet the answer to your question. You don't care about the order of the mice; any order is okay. So you have to consider other possibilities, such as the possibility that the first mouse is in the $80$–$100$ gram range, the next two mice have mass $>100$ grams, the two after that have mass $<80$ grams, and the final three again have mass in the $80$–$100$ gram range, giving you the term

\begin{align*}\tag{$\ast\ast$}(.34)(.50)^2 (.16)^2(.34)^3.\end{align*}

There are many different possible orders, corresponding to mutually exclusive events, so to get the total probability that one of them happens you have to add them up: $(\ast) + (\ast\ast) + \cdots$. Fortunately this is not as much work as it seems like, because the order doesn't matter for multiplication. We just have to figure out how many copies of the term $(\ast)$ to add up.

In other words, we have to figure out how many ways to arrange eight mice in a line, where four are in a certain weight range (call it "A"), two are in a second weight range B, and two are in a third weight range C, and we don't distinguish between different mice in the same weight range. So we can think of this as counting the number of words like AAAABBCC, ABBCCAAA, etc. The number of such words is given by $$N = \frac{8!}{4!\,2! \,2!}.$$

This is because the $8!$ counts the number of ways to arrange 8 letters, and we divide by $4!$, $2!$, and $2!$ because we don't care how the four A's are arranged among themselves, and similarly for the two B's and the two C's. Now to get the final answer we multiply this number $N$ by the probability $(\ast)$.

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