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i would like to exactly what is asked in this problem:

The probability that A can solve the problem is $1/4$ and that B can solve it is $1/3$. If both of them try, what is the probability that problem will be solved.

$A. 1/4$

$B. 7/12$

$C. 1/3$

$D. 1/2$

$E. 1/12$

i will say what is a point of my confusion,when we are trying to find probability of two independent events,we are multiplying probability of each other to get probability that both event occur .now first it is what i have tried and got $1/12$,but it shows me that it is not correct,now i have this question and please could help me,how can i translate if both of them will try into probability language?

$P(A or B)$ is not correct because it means that one or second will solve,so it means that $P(A and B)$,but they are independent are this would not be product of their probabilities?let us denote probability that $A$ will solve by $P(A)$, and probability that $B$ will solve by $P(B)$



now what would be $P(A and B)$?would not it be $1/4*1/3$?

or i am calculating wrongly and it would be $P(A or B)$

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It's 1 minus probability of "not A and not B". – Gerry Myerson Aug 1 '13 at 9:29
but in terms of $A$ and $B$ itself? – dato datuashvili Aug 1 '13 at 9:30
probability of not-A is 1 minus probability of A. – Gerry Myerson Aug 1 '13 at 9:32
@GerryMyerson this one i know,i am asking different thing – dato datuashvili Aug 1 '13 at 9:45
OK: what thing is it that you are asking? – Gerry Myerson Aug 1 '13 at 9:47
up vote 1 down vote accepted

You're trying to find the probability that either A or B solves the question. An easier way to think about it is this - the only way that the problem does not get solved is if both cannot solve it. We can find the probability that A and B cannot solve it, and subtract this from one.

As you mentioned, the probability of A not solving AND B not solving is simply the product of these. Hence, we find the probability that A cannot solve, which is $1 - \frac 14 = \frac 34$. Similarly, we find the probability that B cannot solv, which is $1 - \frac13 = \frac 23$. Multiplying, we have $ 1- \frac34 \times \frac 23$. This gives us an answer of $ 1- \frac12 = \frac 12$.

Hope that helps. Cheers!

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i see,i understood thanks very much – dato datuashvili Aug 1 '13 at 9:37
only one thing is why i can't use formula for $P(A and B)$ – dato datuashvili Aug 1 '13 at 9:39
for such problem,we are saying that it is equal to 1- neither A nor B,but why it could not be simply product? – dato datuashvili Aug 1 '13 at 9:44
Because the product obviously gives the wrong answer. The product is smaller than either of the individual probabilities, whereas the answer must be bigger than either of the individual probabilities. – Gerry Myerson Aug 1 '13 at 9:48
@dato: The question asks for the probability that the problem gets solved. This can happen in any of three ways: A solves it and B does not, B solves it and A does not, or both of them solve it. The formula for P(A and B) only gives you the probability of the third "way". – HJ32 Aug 1 '13 at 10:03

If you want to do it directly (though the proposed way by Gerry is both easier to understand and easier to carry on, imo), you can argue as follows:

The event whose probability we want is "$A$ solves the problem or $B$ solves the problem", and this in numbers is given by "prob. that $A$ solves + prob. that $B$ solves minus the prob. that they both solve


Can you see why we have to substract the probability that they both solve the problem?

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but last one probability that they both solve it is asked exactly,i am confused – dato datuashvili Aug 1 '13 at 9:54
@dato, the fact that both $\,A,B\,$ solve the problem already is contained or observed in the "either $A$ solves the problem or $B$ solves the problem" part, since "or" in mathematics is inclusive, not exclusive. – DonAntonio Aug 1 '13 at 9:57
i see ,so in other word,probability that if they both will try,problem would be solved is the same as either A or B will solve? – dato datuashvili Aug 1 '13 at 10:00
consider that it has answer $1/12$ for confusion – dato datuashvili Aug 1 '13 at 10:01
Yes to your first question, as long as we're clear that saying "either A solves the problem or B solves the problem" contains within it the even "Both A and B solve the problem". – DonAntonio Aug 1 '13 at 10:13

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