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If you want to find the intersection of two dependant events the formula is: P(A and B)= P(A) x P(B|A)

However, what happens if you aren't given P(A and B) as well as P(B|A)? How would you be able to solve that? P(A) x P(B) won't work because that only counts for independent events. Is there a general formula for dependent events?

kjetil b halvorsen
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Lucas
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3 Answers3

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If you are only given $P(A)$ and $P(B)$ and you haven't any information on independence, you can't know $P(A\ and\ B)$.

I'll show that with the same $P(A)$ and $P(B)$ we can get different $P(A\ and\ B)$. For example, suppose we are tossing a coin: $$A=heads$$ $$B=tails$$ $$P(A)=0.5=P(B)$$ $$P(A\ and\ B)=0$$ But for different events with the same probabilities: $$A=heads$$ $$B=not\ tails$$ $$P(A)=0.5=P(B)$$ $$P(A\ and\ B)=0.5$$ However, you can get upper and lower bounds on $P(A\ and\ B)$. $$min(P(A),P(B))\geq P(A\ and\ B)\geq P(A)+P(B)-1$$ Please notice that both bounds only coincide when $P(A)=0=P(B)$ or $P(A)=1=P(B)$.

Furthermore, if some extra information is given you can compute $P(A)$ and $P(B)$. For example, you could be given $P(A\ and\ not\ B)$ or $P(not\ A\ and\ not\ B)$, or you could be told that $A$ and $B$ are independent.

Pere
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    While your answer is correct, I am afraid that your example isn't very clear to illustrate it. – Tim Sep 27 '16 at 10:33
  • @Tim In fact the answer to what the OP asked is just the first sentence. The remaining of the answer are just hints on what the OP might need but doesn't ask and to what future readers might be interested in. I gave the most clear short example I could find but clearity is very subjective and I think the question and the OP would benefit from having another example you might be interested to add. Clear or not, two examples is nearly always clearer and better than just one example. – Pere Sep 27 '16 at 10:50
  • I believe the clause `P(A)+P(B)-1` should read `P(A) + P(B) - P(A) * P(B)` – Chris Sep 15 '20 at 17:27
  • @Chris : you seem to be assuming that the events are independent. Furthermore, please notice that your version of the clause doesn't hold for the coins examples in the answer. – Pere Sep 15 '20 at 19:32
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A Venn diagram addresses this problem.

Two circles are your two marginals, with areas P(A) and P(B). They can overlap to any degree. You cannot deduce their overlap P(A & B) from their individual areas alone.

There is no formula for what you want.

Olexunder
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As noted by Pere, you haven't got enough information. Let me give you an example. To make things simple imagine that we have two events, that the weather is sunny (let's denote it as $A$) and that people eat ice creams (let's denote it as $B$). To make it even simpler, let's say that you calculate empirical probabilities using one-year data about weather and eating ice creams in your local neighborhood. You made observations for $n$ days, among those days there was $n_A$ sunny days and $n_B$ days where people ate ice creams, this gives us

$$ P(A) = \frac{n_A}{n}, \qquad P(B) = \frac{n_B}{n}$$

Does it tell us anything about relation between $A$ and $B$? Well, it doesn't. To know if the two events are related (non-independent) we would need to know how many days there were where people ate ice creams and it was sunny $n_{AB}$, and calculate joint, or conditional probabilities

$$ P(A \cap B) = \frac{n_{AB}}{n}, \qquad P(B\mid A) = \frac{n_{AB}}{n_A}$$

Hopefully, if you know either of those information, you can calculate the another

$$ P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(A \cap B) = P(B \mid A)\,P(A) $$

If you do not know if those two events happen together more, or less often then by chance, how would you know that there is any relation between them? Basically this is what non-dependence is about: that things have different probability of occurring together, then by chance.

Tim
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