Suppose that $10\%$ of all students play a winter sport, $20\%$ play a spring sport, and $5\%$ play both a winter and spring sport. What is the probability that a randomly chosen student plays a winter or spring sport?
[Math] the probability that a randomly chosen student plays a winter or spring sport
probability
Related Solutions
That formula only works if events $A$ (play basketball) and $B$ (play baseball) are independent, but they are not in this case, since out of the $18$ players that play baseball, $13$ play basketball, and hence $P(A|B) = \frac{13}{18} < \frac{22}{30} = P(A)$ (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)
So: the two events are not independent, and so that formula doesn't work.
Fortunately, a formula that does work (always!) is:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Hence:
$$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{22}{30} + \frac{18}{30} - \frac{27}{30} = \frac{13}{30}$$
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.
Best Answer
Use this:
$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
Where $P(A)$ is the probability for winter / spring; $P(A \cup B)$ is the union, namely the probability for both winter and spring and $P(A \cap B)$ is the probability for the OR events.
So:
$$P(A\cap B) = 10 + 20 - 5 = 25\%$$