# How do you find the intersection of two dependent events when you don't have the conditional probability?

by Lucas   Last Updated September 27, 2016 08:08 AM

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?

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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 $P(not\ A\ and\ not\ B)$.

Pere
September 27, 2016 10:26 AM

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 joined, 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
September 27, 2016 10:51 AM