# 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∩B_1)=0.075, P(A_1∩B_2)=0.245. Find P(A_2∩B_3).

Freddy Friedman 2022-07-23 Answered
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, $P\left({A}_{1}\right)=0.4$, $P\left({B}_{2}\right)=0.35$, $P\left({B}_{1}\right)=0.25$ and $P\left({A}_{1}\cap {B}_{1}\right)=0.075$, $P\left({A}_{1}\cap {B}_{2}\right)=0.245$. Find $P\left({A}_{1}\cap {B}_{1}\right)=0.075$.
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 $P\left({A}_{2}\cap B3\right)=P\left({A}_{2}\right)\ast P\left(B3\right)$ 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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Sandra Randall
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 $P\left({A}_{2}\right)$ and $P\left({B}_{3}\right)$, you would already know the answer:
$P\left({A}_{2}\cap {B}_{3}\right)=P\left({A}_{2}\right)P\left({B}_{3}\right)$
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:
$P\left({A}_{2}\cap {B}_{3}\right)>P\left({A}_{2}\right)P\left({B}_{3}\right)$
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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