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In a relay race there are five teams $A, \,B, \,C, \,D$ and $E$. What is the probability that $A,\, B$ and $C$ are first three to finish (in any order) (Assume that all finishing orders are equally likely)
$\frac{1}{10}$
$\frac{1}{10}$
$\frac{1}{10}$
$\frac{1}{10}$
Solution
If we consider the sample space consisting of all finishing orders in the first three places, we will have $^{5} P _{3},$ i.e., $, \frac{5 \,!}{(5-3) \,!}$ $=5 \times 4 \times 3=60$ sample points, each with a probability of $\frac{1}{60}$.
$A$, $B$ and $C$ are the first three finishers.
There will be $3 \,!$ arrangements for $A, \,B$ and $C$.
Therefore, the sample points corresponding to this event will be $3 \,!$ in number.
So $P( A , \,B $ and $C$ are first three to finish) $=\frac{3\, !}{60}=\frac{6}{60}=\frac{1}{10}$