A bag holds ten identical balls numbered 1 to 10. Three balls are drawn at random and their numbers recorded. Find (1) the probability that the smallest number is 5, and (2) the probability that the largest number is 5.

Step-by-step solution

1. 1

   Count the sample space. All $3$-element subsets of $\{1,\ldots,10\}$ are equally likely:

   $\binom{10}{3}=\frac{10\cdot 9\cdot 8}{6}=120$

2. 2

   Translate "the smallest is 5" into a constraint. It means $5$ is one of the three numbers and the other two both exceed $5$. The numbers above $5$ are $6,7,8,9,10$ — five of them.

3. 3

   (1) Count and divide.

   $\binom{5}{2}=10\quad\Longrightarrow\quad P(\min=5)=\frac{10}{120}=\frac{1}{12}\approx 0.0833$

4. 4

   Translate "the largest is 5". Now $5$ must be drawn and the other two must be below it, chosen from $1,2,3,4$ — only four candidates.

5. 5

   (2) Count and divide.

   $\binom{4}{2}=6\quad\Longrightarrow\quad P(\max=5)=\frac{6}{120}=\frac{1}{20}=0.05$

   $\boxed{P(\min=5)=\tfrac{1}{12},\qquad P(\max=5)=\tfrac{1}{20}}$

6. 6

   Check the whole distribution of the minimum. For $\min=k$ the count is $\binom{10-k}{2}$, so summing over $k=1,\ldots,8$ gives $\binom92+\binom82+\cdots+\binom22=36+28+21+15+10+6+3+1=120$ ✓, matching the sample space. The asymmetry between the two answers is expected: there are more numbers above $5$ than below it, so $5$ is more likely to be the smallest than the largest.