# Probability (independent?) events

Hi everyone I have a question about the following problem:

Events for a family: $A_1$ = ski, $A_2=$ does not ski, $B_1$ = has children but none in 8-16, $B_2$ = has some children in 8-16, and $B_3$ = has no children. Also, $P(A_1) = 0.4$, $P(B_2) = 0.35$, $P(B_1) = 0.25$ and $P(A_1 \cap B_1) = 0.075$, $P(A_1 \cap B_2) = 0.245$. Find $P(A_2 \cap B_3)$.

Here is my solution:

Since P(A1 and B1) = 0.075, P(A2 and B1) = 0.25-0.075= 0.175. Also since P(A1 and B2) = 0.245, P(A2 and B2) = 0.35 - 0.245 = 0.105. From this we can find P(A2 and B3) which is 0.6-0.175-0.105 = 0.32. But when I use the formula for independent events formula $P(A_2 \cap B3) = P(A_2)*P(B3)$ I get 0.24. Does this mean that the events are not independent? If so, how are these events not independent?

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Yes, it means the events are not independent. If they were, you would could find a simpler solution, requiring less data. For instance, if you knew $P(A_2)$ and $P(B_3)$, you would already know the answer: $$P(A_2 \cap B_3) = P(A_2) P(B_3)$$

It is perfectly legal for events that appear not to have a direct connection to be independent. It just might be that you're looking at a small population (a small village?) where by pure chance you find a lot of families with small children who ski, but few families without children who ski. You could probably think up some "real world" rationalization for a correlation as well (like, people with children have less time and are less likely to ski, or - conversely - have more incentive to ski with their children).

By the way, since you got: $$P(A_2 \cap B_3) > P(A_2) P(B_3)$$ the two events are positively correlated. It means, intuitively, that if you select a random family, once you learn that $A_2$ holds then the probability of $B_3$ increases. Note that being positively correlated is symmetric, and correlation does not imply causation.

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$A_{2}$ and $B_{3}$ are independent iff $P(A_{2}\cap B_{3}) = P(A_{2})*P(B_{3})$, so yes this means they are not independent. If they were, this wouldn't really be a problem... Actually, the definition is a bit more nuanced...even if we show that $P(A_{2}\cap B_{3}) = P(A_{2})*P(B_{3})$, $A_{2}$ and $B_{3}$ are only truly independent if events $A$ and $B$ are independent, which is true iff $P(\bigcap_{i=1}^{n}A_{i}) = \prod_{i=1}^{n}P(A_i)$ for all finite subsets.

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@DilipSarwate: You are able to edit the post and change the notation if you find it so disagreeable. I think a down-vote is very harsh if notation is the only problem. – Fly by Night Sep 9 '12 at 16:27
@FlybyNight I have removed the down vote since the OP has edited his answer. – Dilip Sarwate Sep 9 '12 at 17:35
@DilipSarwate That's very fair of you. Thank you. – Fly by Night Sep 9 '12 at 17:51

Two events $X$ and $Y$ are independent iff $P(X\cap Y)=P(X)\cdot P(Y)$ (cf. e.g. http://en.wikipedia.org/wiki/Independence_%28probability_theory%29 )

According to your calculation $P(A_2\cap B_3)\ne P(A_2)\cdot P(B_3)$, hence "not skiing" and "not having children" are not independent. In fact, we even see that $P(A_2|B_3) = \frac{P(A_2\cap B_3)}{P(B_3)}$ is significantly bigger than $P(A_2)$.

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