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If there is 25% chance for event A to happen and there is a 25% chance for event B to happen...What is the % probability for both event A and B to happen.

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You need additional information to solve this (e.g. if $A$ and $B$ are independent, then the probability you are looking for is 12.5%). If they are not independent you can solve it by knowing the probability that at least one of them occur. –  Stefan Hansen Sep 20 '12 at 16:46
    
Yes they were 2 independent events so not much more additional information. Thank you. –  Alex Sep 20 '12 at 16:48
    
@RossMillikan: Of course, it is me who can't multiply $1/4$ by $1/4$ correctly. Should be 6,25%. –  Stefan Hansen Sep 20 '12 at 16:57
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@Alex "... so not so much more additional information" You are missing the point. The information that the events are independent, even though you don't regard it as adding much to the problem, is crucial in that it enables you to solve the problem. –  Dilip Sarwate Sep 20 '12 at 17:10
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3 Answers 3

If they are not independent all you can say is P(A and B)<= min [(P(A), P(B)]. This will be true even if they are independent. I thought that A and B can be constructed to have a certain type of dependence such that any value less than or equal to

min[(P(A), P(B)] possible making it a tight upper bound. But Dilip pointed out that if you choose A and B so that P(A)+P(B)>1 then P(A and B) must be greater than 0 and certainly such A and B can be found..

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Why $< \min\{P(A), P(B)\}$ instead of $\leq \min\{P(A), P(B)\}$? Also, are you claiming that for arbitrary events $A$ and $B$ (not necessarily those in the OP's question), $P(A\cap B)$ can have any value in the interval $[0, \min\{P(A), P(B)\})$, or $[0, \min\{P(A), P(B)\}]$ ? –  Dilip Sarwate Sep 20 '12 at 20:23
    
Sorry @DilipSarwate you are right equality occurs if A is a subset of B. I was think of general sets but I think it would hold true if we required P(A)=P(B)=1/4 or any othe fixed probability p such that 0<p<1. Of course with the correction to my statement I am asking about the closed interval rather than the half open one. Thank you for catching that. –  Michael Chernick Sep 20 '12 at 20:34
    
For arbitrary events $A$ and $B$, it is not true that $P(A\cap B)$ can take on any value in $[0, \min\{P(A), P(B)\}]$. The upper bound $\min\{P(A), P(B)\}$ is always attainable (when one event is a subset of the other), but the lower bound $0$ is not necessarily attainable. This happens when $P(A)+P(B)$ exceeds $1$ (which is not the case in the OP's question). –  Dilip Sarwate Sep 20 '12 at 21:03
    
@DilipSarwate Thank you for answering my question. It took a little thought for me to see why 0 cannot be obtained for arbitrary A and B. Your logic is quite good. If P(A)+P(B)>1 then P(A U B) = P(A) + P(B) - P(A and B) implying P(A and B)>0. –  Michael Chernick Sep 20 '12 at 23:00
    
In fact, $P(A \cap B) \geq \max\{0, P(A) + P(B) - 1\}$ and the lower bound is positive when $P(A) + P(B) > 1$. –  Dilip Sarwate Sep 21 '12 at 1:19
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If they're independent, their probability should have been P(A)*P(B) = $\frac {1}{16}$

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If they are independent, then the probability of both is the product of the two probabilities.

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