# 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?

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.

$A_{2}$ 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.

• @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. Sep 9, 2012 at 16:27
• @FlybyNight I have removed the down vote since the OP has edited his answer. Sep 9, 2012 at 17:35
• @DilipSarwate That's very fair of you. Thank you. Sep 9, 2012 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 $P(A_2)$.