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Say we have $P(A) = 0.60,\, P(B) = 0.50$ .

Normally to find the probability of at least one happening, we find the probability of neither of them happening:

$$P(A^c \cap B^c) = 0.40 \times 0.50 = 0.20$$

and then subtract it from $1$ ( getting $0.80$). I know that this is incorrect because we are also given that $P(A \cap B) = 0.35$, thus $A,B$ are not independent. How would one go about finding the probability of at least one occurring in this case?

lulu
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Hello
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2 Answers2

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The event $C=$"at least one from A or B happened" is the same as the event $C=A\cup B$.

So.. just use one of the most basic results in probabilities: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.6-0.35=0.75.$$

bluemaster
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  • Thank you it makes sense now. So if I was to have a similar problem: Suppose A and B are connected in parallel. Only one component must work for the system to work. $P(A) = 0.9$ and $P(B) = 0.8$. Calculate the probability that the system works. I assume it's the same idea but because they're independent (given) I can just jump straight to multiplying their complements and subtracting from 1 to get a final answer of $0.98$? – Hello Aug 29 '17 at 23:55
  • @Hello: Sure. Notice that $P(A\cup B)=1-P(\overline{A\cup B})=1-P(\bar A\cap \bar B)$. If $A$ and $B$ are independent, their complements are also independent and, in this case, $P(A\cup B)=1-P(\bar A)P(\bar B)$. But that works only if $A$ and $B$ are independent. – bluemaster Aug 30 '17 at 00:57
  • Got it. So for this specific case (not the original question, the follow-up) since it is specified that they are independent I can safely conclude that the answer is $0.98$? – Hello Aug 30 '17 at 01:34
  • @Hello: Yes the answer in this case is 0.98 that is $P(A\cup B)=1-P(\bar A)P(\bar B)=1-0.1\times 0.2=0.98.$ – bluemaster Aug 30 '17 at 01:42
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Hint: The probability of A happening and B not happening is $0.6-0.35=0.25$.

Paul
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  • I can easily believe that you know how to solve this problem, based on the type of "hint" offered, but I don't read the Question as asking for a hint. So I'd suggest a one-line note like this is better offered as a Comment, and that if you wish to Answer, you'd not be breaching any confidences by giving an explicit solution. – hardmath Aug 29 '17 at 01:45
  • Not a hint, rather an oracle. Useless. – Did Sep 02 '17 at 15:17
  • I do not know what "the other 2" are but frankly, I do not understand the value you put into this one. Could anybody be asking the question and at the same time be able to guess where your $0.35$ is coming from? No way! – Did Sep 02 '17 at 17:36
  • Hmm, yeah I forgot that what you skipped was how to go from $P(A\setminus B)$ to $P(A\cup B)$ (and please, if you add the argument, do it in your answer for the OP's sake, since doing it in a comment for my sake would be quite useless). – Did Sep 02 '17 at 18:25