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In a high school graduating class of 100 students, 54 studied mathematics, 69 studied history, and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that

a). The student took mathematics or history

b). the student did not take either of these subjects

c). the student took history but not mathematics

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    $\begingroup$ Can you draw the Venn diagram for the given information? $\endgroup$ Jun 11, 2014 at 14:53

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If you draw the venn diagram as David suggested, the problem will be much clearer.

If you have 100 students, then

a) 69-35 = 34 students who studied only history. Likewise 54-35 = 19 who studied only math. This gives us 34+19 = 53 students who studied math OR history but not both. Hence the probability of choosing one of these students is 53/100. The question doesn't specify inclusive or exclusive OR. If it is inclusive or, then you have to add the students who took both math and history. This gives you 53 + 35 = 88 students hence a probability of 88/100.

b)The remaining students 100-88 = 12 didnt take any math nor history hence a probability of 12/100

c) From all the history students, you subtract the amount that did both math and history. So 69-35 = 34 students, hence a probability of 34/100

Hope this clears it up!

EDIT: You may want to read up on conditional probability and Bayes theorem, because I assume you're learning these techniques to solve these types of problems. EDIT2: I changed each answer, sorry about misleading you before.

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  • $\begingroup$ thanks for the help, but something is not calculating correctly, because the answers are: a). P(MuH)=88/100 = 22/25 b). P(M'nH') = 12/100 = 3/25; c). P(HuM') = 34/100 = 17/50. I need help solving them to get these answwers $\endgroup$ Jun 12, 2014 at 19:12

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