Found this problem on intmath.com:
The probability that a student passes Mathematics is 2/3 and the probability that he passes English is 4/9. If the probability that he will pass at least one subject is 4/5, what is the probability that he will pass both subjects?
OA is [spoiler]14/45[/spoiler]
What I don't get is why we can't just multiply 2/3 by 4/9?
Probability
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i dont understand what your problem to solve this problem. we dont multiply anything.boysangur wrote:Found this problem on intmath.com:
The probability that a student passes Mathematics is 2/3 and the probability that he passes English is 4/9. If the probability that he will pass at least one subject is 4/5, what is the probability that he will pass both subjects?
OA is [spoiler]14/45[/spoiler]
What I don't get is why we can't just multiply 2/3 by 4/9?
there is a formular: P(a or be)=Pa+Pb-P(a and b) [ pa is probabiliy to pass 1 subject)
for this problem:
4/5=2/3+4/9-P(a and b)
and P(a and bP=2/3+4/9-4/5=14/45
Your reply is correct. I was just wondering why I can't simply multiply the probabilities that he passes the math (2/3) and english (4/9). The problem is asking for the probability of him passing both courses, math AND english, isn't it P(m) x P(e)?diebeatsthegmat wrote:i dont understand what your problem to solve this problem. we dont multiply anything.boysangur wrote:Found this problem on intmath.com:
The probability that a student passes Mathematics is 2/3 and the probability that he passes English is 4/9. If the probability that he will pass at least one subject is 4/5, what is the probability that he will pass both subjects?
OA is [spoiler]14/45[/spoiler]
What I don't get is why we can't just multiply 2/3 by 4/9?
there is a formular: P(a or be)=Pa+Pb-P(a and b) [ pa is probabiliy to pass 1 subject)
for this problem:
4/5=2/3+4/9-P(a and b)
and P(a and bP=2/3+4/9-4/5=14/45
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if you multiply the 2 prob of maths and english then you are taking into account the joint probability as well.(i.e a person could have passed both subjects and this is inlcuded in individual prob).so you are counting it twice once with maths and once with english.
so you need to subtract this prob once from sum of individual prob p(M) +p(E)-p(M and E)
so you need to subtract this prob once from sum of individual prob p(M) +p(E)-p(M and E)