Probability that at least 1 student is taking a language class solution

Question

This is a problem from A First Course in Probability by Ross:

An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that are in both French and German. In addition, there are 2 students taking all 3 classes.

The question I am trying to answer is:

If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

From a previous question, the answer that a randomly chosen student is not in any of the language classes is 1/2. The answer to this question is 149/198, and it is calculated as $$1-\frac{50\choose2}{100\choose 2}=1-\frac{49}{198}=\frac{149}{198}.$$

My question is why can't this be calculated as $$1 - P(\text{Both students not taking language classes}) = 1 - \left( \frac{1}{2} \right)^2 = \frac{3}{4}$$ since the individual probability that a student is not taking a language class is 1/2?

Answer 1

Notice that $\frac{149}{198} = 0.7525... \approx 0.75 = \frac34$, so your answer is not far off!

Both answers initially calculate the probabilty that neither student is taking a language class, this is where the difference comes from.

What your solution ignores is the fact that after you assume that the first randomly selected student is not taking a language class, the probability that that is also true for the second student changes slightly, from $50$ out of $100$ ($=\frac12$) to $49$ (remaining students not taking a language class) out of $99$ (remaining students to choose from), which is slightly less than $\frac12$. If you account for that, your answer becomes

$$1-\frac12\frac{49}{99} =1- \frac{49}{198} = \frac{149}{198},$$

matching the answer given.

If you are not convinced why the second probability would change, think about what would happen if you continue selecting students. You don't select 2 students, but 3, 4,... 51 students. If the probability of the next student not taking a language class would always stay $\frac12$, the probability for $51$ students all not taking a language class would be $\frac1{2^{51}}$. Which is certainly very small, but still positive. But in reality, selecting 51 students from only 50 (those that don't take language classes) is impossible, not just highly unlikely.

Your answer is a good approximation for selecting a 'small' number of students from a 'big' group, where selecting a few students doesn't change the probability much $\frac{49}{99} \approx \frac12$. Where exactly the boundary is between 'small' and 'no longer small' is a topic of advanced statistics/probability theory.

Answer 2

$P$(Both students not taking language classes) ≠ $\left( \frac{1}{2} \right)^2$

Let event A = first student is not taking any language class and event B = second student is not taking any language class

Note that the two events are not independent. When event A has occurred, there is one less person we need to consider in our sample space for event B to occur. i.e.,

$=> P$(Both students not taking language classes) = $P(A) \cdot P(B) = \frac12 \cdot \frac{49}{99} = \frac{49}{198}$

Answer 3

The anwsers in this case are similar. As both probabilities can be rounded off to 3/4. Yet there is one important difference; which is how you deal with replacement.

To be sure; we know 50 students are taking a class and 50 students are not taking a class. In the calculation you stated we assume that the draws are independent. Hence you would get 1 - (1/2 * 1/2) = 3/4. As you stated.

However the combination approach assumes that we pick students without replacement, this means that once a student is chosen we can't put them into the pool again of students that can be chosen. Hence we can't simply multiply the same probability.

Using combinations in this case does still give a similar result yet the underlying assumption is different, and this matters as the number of students selected increases;

We can simplify and re-write your combination equation as;

1 - ((50 * 49 / 2) / (100 * 99 / 2) = 0.753

Note. what we do here is the same as saying; We pick one student out of 50, and then one out of 49. And similarly we pick one student out of a 100, and then one out of 99.

In words; total probability - (ways to select two students without replacement of 50 who did not take a class / total ways to pick two students without replacement out of 100 students) = 0.753.

Which one is better?

It depends on your assumption. In this case the anwser which was given assumes that both students are unique, meaning different. And that we do not have replacement.

Answer 4

As you say, it was previously computed that $2$ students take no language class, and P( a student takes no language class) =$\frac12$,

then we could have computed
P(at least one student takes a language class)
$= 1 - (\frac12)^2 = \frac34$

which would have been close to $\frac{149}{198}$ but not exactly equal to it.

This has to do with the fact that you are selecting without replacement You can see it more clearly by directly multiplying probabilities.The probability from selection to selection will not be constant, so you can't use the special multiplication law, and need to use the general multiplication law Do look up the multiplication laws

Thus P(at least one student takes a language course)

$ = 1 - \frac{50}{100}\frac{49}{99} = \frac{149}{198}$