The question is

I have a class of 10 students. The class consists of 4 statistics majors, 3 philosophy majors, and 3 sociology majors.

I choose three students at random, without replacement. (Order doesn’t matter.)

Let event A be "All three students I choose are statistics majors."

Let event B be "At least one student I choose is a philosophy major."

For P(A) I did $$4/10 \cdot 3/9 \cdot 2/8 = .033$$

For P(B) I thought it would be $$(7!\cdot 3!)/(10!) = .833$$ But that is not correct. I am a little confused on where I am going wrong, I thought it would be number of events that satisfy B/number of equally likely events some help in the right direction is all im asking for, thank you in advance.

  • $\begingroup$ Notice that the "complement" of the event B is equivalent to "No student I choose is from a philosophy major", can you calculate the corresponding probability now? $\endgroup$
    – Fei Cao
    Jan 31, 2021 at 23:44
  • $\begingroup$ How did you get the quantity $\frac{7!3!}{10!}$? What were "number of events that satisfy B" and "number of equally likely events" in your approach? $\endgroup$
    – Clement C.
    Feb 1, 2021 at 0:00
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$
    – Clement C.
    Feb 4, 2021 at 23:14

3 Answers 3


So, to use your approach: for the first, you have indeed $$\Pr[A] = \frac{\binom{4}{3}}{\binom{10}{3}} = \frac{1}{30} $$ since you have $\binom{4}{3}$ events where the 3 students you choose belong to the 4 statistics majors, and a total of $\binom{10}{3}$ events (choose $3$ students out of 10).

For the second, $\Pr[B]$, this is not the fastest way, but you can do the same. The number of events where at least one student is a philosophy major is $$ \binom{3}{1}\binom{7}{2} + \binom{3}{2}\binom{7}{1} + \binom{3}{3}\binom{7}{0} = 85 $$ since you choose either exactly 1, exactly 2, or exactly 3 of your 3 picks to be philosophy majors. In the denominator, of course, you'll still have $\binom{10}{3}$ events.

A faster way is to note that the complement of $B$, which has probability $1-\Pr[B]$, is "none of the students I chose is a philosophy major," and the number of events satisfying this is more simply $ \binom{7}{3} $. Computing the probability of the complement is much easier.


Obviously the easiest approach is $1-P(\text {no philosophy majors})=1-\frac 7 {10} \frac 6 9\frac 5 8$

If you were to do $\frac{\text{number of groups with at least one philosophy major}}{\text{number of groups}}$, you'd have to calculate the number of groups all of which have at least one philosophy major. Notice that this is equivalent to $\frac{\text{number of groups}-\text{number of groups without a philosophy major}}{\text{number of groups}}=\frac{{10 \choose 3}-{7 \choose 3}}{{10 \choose 3}}$. Notice that working with groups is the same as working with probabilities, as this simplifies to $1-\frac{{7 \choose 3}}{{10 \choose 3}}=1-\frac{7 *6 *5}{10*9*8}$


Comment: @ ClementC's answer (+1) is correct, comprehensive, and to the point. Also, both parts can be solved as hypergeometric probability problems, and both answers can be easily approximated by simulation.

Using the hypergeometric PDF function phyper in R: (Exact answers to the stated number of decimal places.)


dhyper(3, 4,6, 3)
[1] 0.03333333


1 - dhyper(0, 3,7, 3)
[1] 0.7083333

Simulation with a million iterations can be expected to give 2 or 3 place accuracy.

# a
class = c(1,1,1,1, 2,2,2, 3,3,3)
nr.1 = replicate(10^6, sum(sample(class, 3)==1))
mean(nr.1 == 3)
[1] 0.03342

# b
nr.3 = replicate(10^6, sum(sample(class, 3)==3))
mean(nr.3 >= 1)
[1] 0.708252

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