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Re: The probability that a student will pass in Math is 3/5 and [#permalink]
what about people who will pass neither one? does the problem exclude them?
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Re: The probability that a student will pass in Math is 3/5 and [#permalink]
AJB77 wrote:
Folaa3 wrote:
7/8

but hsouldn't the probability of passing both math and englush be 1/5 instead of 1/8 ?
Assume that there are 120 students (120 is the LCM of 3,5 and 8)

No of students who passed Math = (3/5)*120 = 72
No of students who passed English = (1/3)*120 = 40
No of students who passed both, which by the way is not calculable from the above = (1/8)*120 = 15
So, the number of students who passed either Math or English or both = 72+40 - 15 = 97
QED.


i see what you are saying but my point is, if the question stem says 3/5 passed maths and 1/3 passed english,
then the prob of those who pass both math and english should be
1/3 * 3/5 = 3/15 = 1/5
Unless the stem is implying information that is not given. How can the prob of both be 1/8.
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Re: The probability that a student will pass in Math is 3/5 and [#permalink]
Folaa3 wrote:
i see what you are saying but my point is, if the question stem says 3/5 passed maths and 1/3 passed english,
then the prob of those who pass both math and english should be
1/3 * 3/5 = 3/15 = 1/5
Unless the stem is implying information that is not given. How can the prob of both be 1/8.


3/5 passed Math could mean that all of the 3/5 passed English as well or none of them passed English.

Let me give you an example:

Let's say we have 6 people, 1 passed Math alone, 1 passed English alone, 2 passed both and 2 failed both

Now if I ask you, How many students passed Math? Will you say 1 or 3?

Similarly, how many students passed English?

Think about it.
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Re: The probability that a student will pass in Math is 3/5 and [#permalink]
Folaa3, 7/8 is the possibility that the student fail at least one subject (vs pass at least one subject). (Also, there is only one student in this case, although not important.)

Passing English test and Math test are not independent, you can't simply multiply the possibilities together and say P(A&B)=P(A)*P(B).



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