$30\%$ students have glasses. $20\%$ of students with glasses play. $60\%$ of students without glasses play. Probability, student without glass plays.

Question

Problem: In a school, $30\%$ of students have glasses. $20\%$ of students with glasses play sports. $60\%$ of students without glasses play sports. If we randomly choose a student, find probability that a student without glasses (chosen randomly) plays a sport.

My Approach

• $P(O)=0.3$ is the probability that a student has glasses.
• $P(S|O) = 0.2$ is the probability that a student with glasses plays sport (P of S given O).
• $P(S|O^c)=0.6$, where $O^c$ is {Sample Space - $O$}. The probability that a student without glasses plays sport $(P(S\text{ given }O^c)$.

I have to find $P(O^c|S)$ (the opposite of $P(S|O^c)$, meaning "a student plays sport without glasses").

I know that $80\%$ of students play a sport, and that $70\%$ of students don't have glasses. So in order to solve the problem I have to do $$P(O^c|S) = \frac{0.8 \times 0.7}{0.8} = 0.7 $$ but given correct solution is $0.87$ $(0.7\neq0.87)$.

EDIT: I think the problem is in this assumption: I've assumed that $0.8$ and $0.7$ were two independent probabilities, therefore I've multiplied $0.8$ with $0.7$. If this is not correct, then I don't know how to find probability of intersection.

Answer 1

$$\mathbb{P}(O^c|S)=\frac{\mathbb{P}(O^c\cap S)}{\mathbb{P}(S)}=\frac{\mathbb{P}(S|O^c)\mathbb{P}(O^c)}{\mathbb{P}(S|O^c)\mathbb{P}(O^c)+\mathbb{P}(S|O)\mathbb{P}(O)}=\frac{0.6*0.7}{0.6*0.7+0.2*0.3}=0.875$$

Answer 2

in a school, 30% of students have glasses. 20% of students with glasses play sports. 60% of students without glasses play sports. if we choose randomly a student, find probability that a student without glasses (chosen randomly) plays a sport.

I disagree with the OP's (i.e. original poster's) interpretation of the problem. This implies that I also disagree with the analysis in the answer of SacAndSac. It also implies that I disagree with the answer that was given as correct, namely $0.875$.

The reason that I disagree is because of how the problem is worded. The statement:

...find probability that a student without glasses (chosen randomly) plays a sport.

is interpreted by me to indicate that you are to confine your focus to only those students that are without glasses, and to then determine what fraction of them play a sport.

Certainly, there are problems with my interpretation:

• For one thing, my interpretation implies that the problem is trivial, because you are already given that

60% of students without glasses play sports.

This implies that the information that I claim is being asked for has already been directly handed to you.

• For another thing, if you reverse engineer on the answer of $0.875$, it seems clear that the intent is to assume that your focus is confined to those people who play sports, and to determine what fraction of them don't wear glasses.

The real difficulty here is that although sometimes the OP will misinterpret a problem, I don't think that is the case here. The impression that I got is that the OP faithfully quoted the problem composer's wording.

So what you have is a problem whose wording makes the problem a cross between trivial and nonsensical. So the MathSE reviewers and the OP are reversing the interpretation into a sensible problem.

That is all fine, except that that isn't the problem that the problem composer worded, assuming that the OP faithfully presented the problem's wording.

Answer 3

in a school, 30% of students have glasses. 20% of students with glasses play sports. 60% of students without glasses play sports. if we choose randomly a student, find probability that a student without glasses (chosen randomly) plays a sport.

It seems like the first part means choose any student at random from all students, but the second part seems to mean we are picking a random student without glasses. That can't be the case because we are given that 60% of students without glasses play a sport so that would give us the answer of 0.6 directly. Given that you seem to think the correct answer is 0.87, I'm going to assume we are choosing a random student from the pool of students that play sports and trying to find the chance that the random sport-playing student does not wear glasses.

Given that 30% of students have glasses, 70% of students do not. But you are incorrect thinking that 80% of students play sports. 20% of group A play sports and 60% of group B, so the answer must be between 20% and 60%.

With the nice percentage numbers, let's assume that we have 100 students. They fall into 4 groups:

1. 30% (glasses) * 20% (glasses and sports) = 6 students have glasses and play sports
2. 30% (glasses) * 80% (glasses and no sports) = 24 students have glasses and don't play sports
3. 70% (no glasses) * 60% (no glasses and sports) = 42 students without glasses play sports
4. 70% (no glasses) * 40% (no glasses and no sports) = 28 students without glasses and don't play sports

We're sampling from groups 1 and 3 that play sports (a total of 48 or 48% of students play sports), so we have 6 students with glasses and 42 students without glasses. That means the chances of a random students that plays sports to not have glasses is 42/(42 + 6) which simplifies to 7/8 or 0.875.