I was studying for some quizzes when I was faced with this problem. It goes like this:
In the senior year of a high school graduating class of $100$ students, $42$ studied Mathematics, $68$ studied Psychology, $54$ studied History,
$22$ studied both Mathematics and History, $25$ studied both Mathematics and Psychology, $7$ studied History but neither Mathematics nor Psychology,
$10$ studied all three subjects, and $8$ did not take any of three. If a student is selected at random, find the probability that(a) a person enrolled in psychology takes all three subjects;
(b) a person not taking psychology is taking both history and mathematics.
My work:
I'm going to try to answer (a).
I guess this particular problem is a conditional probability, because I think that the preceding probabilities will affect the new probabilities. To be specific, I'm looking for the probability that a person enrolled in Psychology takes all three subjects (I denote this as
$P(Takes\space all\space three\space subjects|Psychology)$) or probability that the person will take all three subjects, given that the person is enrolled in Psychology.
Then, I believe that I can get $P(Takes\space all\space three\space subjects|Psychology)$ by using the equation
$$P(Takes\space all\space three\space subjects|Psychology) = \frac{P(Psychology \cap Takes\space all\space three\space subjects)}{P(Psychology)}$$
The expression $P(Psychology \cap Takes\space all\space three\space subjects)$ means probability that a person choosen is enrolled in Psychology and takes all three subjects. I don't have that value though.
The expression $P(Psychology)$ means probability that a person chosen is enrolled in Psychology. I believe that this probability would be $\frac{68}{100}$.
Therefore…..I don't know the answer.
I'm going to try to answer (b).
I guess this particular problem is a conditional probability either, because I think that the preceding probabilities will affect the
new probabilities. To be specific, I'm looking for the probability that a person not enrolled in Psychology is taking both History and Mathematics.
(I denote this as $P(History \cap Mathematics|\neg Psychology)$)
Now….this expression $P(History \cap Mathematics|\neg Psychology)$, when I'm going to evaluate it, I don't where to start. Perhaps it is wrong all along, but that is the expression that will accurately describe the second problem.
How do you answer the above question?
Best Answer
I would have commented but I don't have privilege to comment.
For part(a) I agree with your approach. $P(\mathrm{Psychology}\cap\mathrm{Takes \;all\; three\; subjects}) = P(\mathrm{Takes\; all\; three\; subjects})$ since you are taking intersection and psychology is included in three subjects. and probability of taking all three subjects is given in question that is $10/100$.
For part(b) $P(\mathrm{History}\cap\mathrm{Mathematics}|\neg\mathrm{Psychology}) = P(\mathrm{History}\cap\mathrm{Mathematics}\cap\neg\mathrm{Psychology})/P(\neg\mathrm{Psychology})$
You can find the probability of $\neg \mathrm{Psychology}$ by simply subtracting probability of psychology from 1. you can also expand the numerator as $P(\mathrm{History}\cap\mathrm{Mathematics})P(\neg\mathrm{Psychology}|\mathrm{History}\cap\mathrm{Mathematics})$ now $P(\mathrm{History}\cap\mathrm{Mathematics})$ is given in question that is $22/100$. Out of these $22$ students there are $10$ such students that are enrolled in all three subjects so $22- 10 = 12$ students are such students that are enrolled in maths and history and are not taking psychology. hence $P(\neg \mathrm{Psychology}|\mathrm{History}\cap\mathrm{Mathematics}) = 12/32$
You can do rest of calculations by yourself by putting the values. Please consider my solution as a comment.