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At a certain university, 20% of students only major in literature, 15% of students only major in history, and 8% of students major in both literature and history. The literature-only majors spend a semester abroad with probability 0.32, the history-only majors spend a semester abroad with probability 0.49, and the literature-history double-majors study abroad with probability 0.93. Students of other majors study abroad with probability 0.04. If a student from this university spends a semester studying abroad, what is the probability he or she is a double-major in literature and history?

This is what I have so far:

Let the event that a student goes abroad be = A,

for student major in literature = Sl; history = S2, and literature-history double-majors = S12 and others = So

Probabilty for other major student will be => So=1- 0.2 - 0.15 - 0.08 = 0.57

Hence, probabilty of going abroad P(A)=P(Sl) * P(A|Sl) + P(S2) * P(A|S2)P(S12) * P(A|S12) + P(So)*P(A|S1o)

= 0.2 * 0.32 + 0.15 * 0.49 + 0.08 * 0.93 + 0.57 * 0.04 = 0.2347

Hence, the probabilty that a student is a double-major in literature and history, given that he goes abroad =P(S12|A) = P(S12)*P(A|S12)/P(A)

= 0.08 * 0.93/ 0.2347 = 0.31

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  • $\begingroup$ You haven't shown what you've tried to do, or even your thoughts about it. We can't tell what you need help with. $\endgroup$ Jan 29, 2017 at 7:24
  • $\begingroup$ I have edited the question and added what I did. $\endgroup$
    – michael m
    Jan 29, 2017 at 7:42

3 Answers 3

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is the answer 0.317?

First I went to find what is the probability that someone goes abroad.

P(Literature major+abroad) = 0.2 x 0.32 = 0.064

P(History major+abroad) = 0.15 x 0.49 = 0.0735

P((Literature/History major+abroad) = 0.08 x 0.93 = 0.0744

P(other majors+abroad) = (1-0.2-0.15-0.08) x 0.04 = 0.0228

Therefore, the probability that someone from that university goes abroad = 0.064+0.0735+0.0744+0.0228 = 0.2347.

Hence, probability of a double major going abroad given that someone went overseas = 0.0744/0.2347 = 0.317

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  • $\begingroup$ Got the same answer. Thank you. $\endgroup$
    – michael m
    Jan 29, 2017 at 9:01
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Draw a hundred boxes, each of which has one hundred sub-boxes.

Outline the big boxes in colors as follows:

  • Outline twenty in blue (literature);
  • Outline fifteen in red (history);
  • Outline eight in purple;
  • Outline the rest in green.

In each blue-outlined box, fill in 32 sub-boxes.

In each red-outlined box, fill in 49 sub-boxes.

In each purple-outlined box, fill in 93 sub-boxes.

In each green-outlined box, fill in 4 sub-boxes.

Count all the filled-in sub-boxes.

What fraction of them are in purple-outlined big boxes?


You can do this faster with multiplication and algebraic symbols...maybe.

If you don't understand what you're doing, you won't get a correct answer faster. In that case, get some big pieces of paper and start drawing.


Even without actually drawing anything, I hope this explanation will help you visualize and understand the question better.

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  • $\begingroup$ I think I got it. I need to used The Law of Total Probability in conjunction with Bayes’ Rule, since we have the conditional probabilities of an event and need the (non-conditional) probability of the event. Thank you. $\endgroup$
    – michael m
    Jan 29, 2017 at 8:59
  • $\begingroup$ @michaelm, great. Please take a look at What should I do when someone answers my question? On this site if your question has been fully answered, you can "accept" one of the answers you got by clicking the green checkmark to the left. (See link for more details.) $\endgroup$
    – Wildcard
    Jan 29, 2017 at 10:38
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Your symbols (Sl, S1, S12, So) are chosen poorly, but you have the right approach.

Try to select a symbol set that is consistent and, preferably, self-commenting.   It makes reading (and revision) much clearer.

Subscripts should really only be used when the variables are enumerated, or if variables really need to be collated by type.   Here we can use a different letter for each category of study as the only other event is whether the student goes abroad.

At a certain university, 20% of students only major in literature, 15% of students only major in history, and 8% of students major in both literature and history.

The disjoint events can be represented as $L$ literature-only, $H$ history-only, $D$ double-major, $O$ other-major; a nice meaningful set of characters with no excessive subscripts (though we could have chosen $M$ for the later if concerned about confusion with $0$).

You are given $\mathsf P(L)=0.20, \mathsf P(H)=0.15, \mathsf P(D)=0.08$ and so, $\therefore ~\mathsf P(O)=0.37$

(so far so good)

The literature-only majors spend a semester abroad with probability 0.32, the history-only majors spend a semester abroad with probability 0.49, and the literature-history double-majors study abroad with probability 0.93. Students of other majors study abroad with probability 0.04.

Adding the event $A$ for study abroad, we have:

$\mathsf P(A\mid L)=0.32, \mathsf P(A\mid H)=0.49, \mathsf P(A\mid D)=0.93, \mathsf P(A\mid O)=0.04$

If a student from this university spends a semester studying abroad, what is the probability he or she is a double-major in literature and history?

It is an application of Bayes' Rule and the Law of Total Probability.

$$\mathsf P(D\mid A) = \dfrac{\mathsf P(A\mid D)~\mathsf P(D)}{\mathsf P(A\mid L)~\mathsf P(L)+\mathsf P(A\mid H)~\mathsf P(H)+\mathsf P(A\mid D)~\mathsf P(D)+\mathsf P(A\mid O)~\mathsf P(O)}$$

Which is what you were doing.   Everything else is just ensuring you substituted the right values and crunched the numbers (which looks okay).

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