Adding probabilites to equal 1

Question

I have the following problem:

A farmer has 5 sheep and 4 lambs. He wants to shear 3 of them, picked at random. What are the probabilities that in his selection there are 0, 1, 2 and 3 lambs?

Now, I know that the sum of the probabilities have to add up to 1, because in all of the possible combinations there will necessarily be either 0, 1, 2 or 3 lambs. However, when I add up my probabilities, they do not add up to 1. Here is what I'm doing:

$ 3C3\times\frac{5C3}{8C3} + 3C1\times\frac{4C1}{8C3} + 3C2\times\frac{4C2}{8C3} + 3C3\times\frac{4C3}{8C3}$

Essentially, I've analyzed each individual case - when there are 0 lambs (or conversely, 3 sheep), 1 lamb, 2 lambs and finally three lambs. I've then multiplied each case by the number of ways in which they can be ordered, i.e. 1 lamb can be ordered $3C1 = 3$ ways. However, this all adds up to $\frac{11}{14}$, not to 1.

Can someone please orient me on what I'm doing wrong? Thanks!

Answer 1

You're doing hypergeometrical probability here (that's just the name of any picking process like yours; it's just as easily done with red and blue balls from an urn, or functioning and non-functioning light bukllbs from a box, or anything else). This means that the probability of choosing $n$ lambs is $$ \frac{\binom{4}{n}\binom{5}{3-n}}{\binom{9}{3}} $$ You get this by applying the basic probability formula $\frac{\text{good outcomes}}{\text{possible outcomes}}$. There are $\binom{9}{3}$ ways the farmer can choose what animals to shear, there are $\binom4n$ ways to pick out the four lambs, and there are $\binom5{3-n}$ ways to pick out the remaining sheep.

If you add together all of those for $n = 0, 1, 2, 3$, you will get $1$.