Ten students, $A, B, \ldots,$ are in a class. If a committee of 3 is chosen at random from the class, find the probability that $A$ is chosen.

Question

Ten students, $A, B, \ldots,$ are in a class. If a committee of 3 is chosen at random from the class, find the probability that $A$ belongs to the committee.

My attempted solution:

Since $A$ can be chosen in only $1$ way from $10$, and rest of the two in $9$ and $8$ respectively,

$P(A$ belongs to the committee$) = \frac{1 \cdot 9 \cdot 8}{^{10}C_3} = \frac{72}{120} = \frac{3}{5}$

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But, the correct answer is $\frac{3}{10}$.

What am I missing?

Answer 1

In the numerator, you are counting all permutations of the two members other than $A$, but in the denominator you count only combinations.

In other words, you have counted $ABC$ and $ACB$ as two different outcomes in the numerator, but you counted outcomes in the denominator as if these were just one outcome.

If you use combinations in the denominator, the numerator should be $1\cdot \binom92.$ You get that by selecting $A$ in one way, and then selecting two more members out of $9$ possible members.

Answer 2

You're missing a factor of $2$, since the order in which the other two members are chosen from $9$ and then $8$ is irrelevant.