Lets suppose ten horses are participating in a race and each horse has equal chance
of winning the race. I am required to find the following:
(a) the probability that horse A wins the race followed by horse B.
(b) the probability that horse C becomes either first or second in the race.
I know there are $10 \cdot 9 \cdot 8 $ ways of having first, second or third.
Since each horse has an equal chance of winning, each has probability of 1/10. Would I be right in saying that the probability that A wins followed by B is $\frac{1}{10} \cdot \frac{1}{10} $?
Is it okay if I do this for (b)? $\frac{1}{10} +\frac{1}{10} $?
Best Answer
You can double-check yourself by counting outcomes. There are $10!$ possible finish orders, and it’s implied that they’re all equally likely. If $A$ wins and $B$ finishes second, there are $8!$ equally likely possible orders for the remaining $8$ horses. Thus, there are $8!$ outcomes in which $A$ wins and $B$ comes second out of a total of $10!$ possible outcomes, for a probability of $$\frac{8!}{10!}=\frac{8!}{10\cdot9\cdot8!}=\frac1{10\cdot9}=\frac1{90}\;,$$ just as Andrew Salmon argued directly from the probabilities.
For the second question, if $C$ finishes first, there are $9!$ possible finish orders for the other $9$ horses. The same is true if $C$ finishes second. Since $C$ can’t finish both first and second, these $9!+9!$ outcomes are all distinct, and the probability that $C$ finishes first or second is therefore $$\frac{9!+9!}{10!}=\frac{2\cdot9!}{10\cdot9!}=\frac2{10}=\frac15\;,$$ again just as Andrew argued directly from the probabilities.