# Alternative approach to solving probability problem

Problem
An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. These 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 French and German, and 6 that are in both Spanish and German, In addition, there are 2 students taking all 3 classes. Let S be the event that a student is in Spanish, let F be the event that a student is in French and let G be the event that a student is in German. If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

My solution

Total number of students taking at least one class $=28+26+16-12-6-4+2=50$ Probability that one student out of two is taking class $= \frac{50}{100}\times \frac{50}{100} = \frac{1}{4}$
Probability that both students are taking class $= \frac{50}{100}\times \frac{50}{100}=\frac{1}{4}$
Total probability $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}=0.5$

Given solution

P(at least 1 is taking a language class) = 1 − P(neither is taking a language class) Since there are 100 students in the school and P(does not take a language class) = 0.5, there are 50 students that are not taking any language classes. We want to find the probability that we choose 2 of these students out of all possible choices of two students. That is, $$\frac{\binom{50}{2}}{\binom{100}{2}}$$ Then the probability that at least one of the two students is taking a language class is $$1-\frac{\binom{50}{2}}{\binom{100}{2}}\approx 0.753$$

So where did I make mistake? Should I count "one student of two is taking class" twice? As their are two students, say A and B. Once for A taking class and again for B is taking class. This will make total probability $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}=0.75$. Is it so? Or my approach completely wrong? Also the given answer is 0.753. There is still the difference of 0.003 between 0.753 and 0.75.

The difference comes from the fact that the events are not independent. For example, if you want to calculate that if you pick two students then the probability that neither one of them takes a class is equal to $$\frac{50}{100}\cdot \frac{49}{99}$$ and not equal to $$\frac{50}{100}\cdot\frac{50}{100}.$$

I hope this helps!

• ohhh it seems to be $\frac{50}{100} \times \frac{50}{99} \times 2+ \frac{50}{100} \times \frac{49}{99}=0.752525$, right? – anir Jun 3 '18 at 19:40
• Yes, that it is indeed the case. – Stan Tendijck Jun 3 '18 at 19:45

Since the events "taking a class" are NOT independent among students, by inclusion exclusion principle, the union is evaluated as (bare in mind you're selecting students): $$P(\geq1\text{ of them is taking a class})=P(S_1 \text{ taking})+P(S_2 \text{ taking})-P(S_1 \text{ and } S_2 \text{ taking})=\frac{50}{100}+\frac{50}{100}-\frac{50\times49}{100\times99}$$

So the error occurs when "Probability that both students are taking class $= \frac{50}{100}\times \frac{50}{100}=\frac{1}{4}$".

• So, they are dependent. – Stan Tendijck Jun 3 '18 at 19:46
• Stan, thank you, I typed something wrong. – poyea Jun 3 '18 at 19:47