Little homework help needed

[Kromey]

Hi, I need help with the last question of a homework problem.

I am given: P(A1) = 0.23, P(A2) = 0.26, P(A3) = 0.29, P(A1 ∩ A2) = 0.08, P(A1 ∩ A3) = 0.07, P(A2 ∩ A3) = 0.05, P(A1 ∩ A2 ∩ A3) = 0.01.

I need to find: P(A1 ∩ A2 ∩ A3 | A1 ∪ A2 ∪ A3).

I know:

P(A1 ∪ A2 ∪ A3) =

P(A1) + P(A2) + P(A3) − P(A1 ∩ A2) − P(A1 ∩ A3) − P(A2 ∩ A3) + P(A1 ∩ A2 ∩ A3)

= .23 + .26 + .29 - .08 - .07 - .05 + .01 = .59

and, it is given:

P(A1 ∩ A2 ∩ A3) = 0.01, but I do not know how to combine these for P(A1 ∩ A2 ∩ A3 | A1 ∪ A2 ∪ A3). Can someone help please?

I know P(A | B) = P(A ∩ B) / P(B), but if I extrapolate that with the above, more complex formulas, it yields a completely incorrect answer.

 

[BGM]

Yes you need to use formula from the definition of conditional probability, and you already calculated the denominator. The numerator is easy to be simplified if you can recognize one set is the subset of another.

i.e. For any 2 sets \( E, F \),
\( E \cap F \subseteq E \subseteq E \cup F \)

and subsequently it can help you to simplify the intersection in numerator.

 

[Sugata Das]

In fact, with the formula P(A|B)=P(A∩B)/P(B), you should have the numerator like this

P[(A1∩A2∩A3) ∩ (A1UA2UA3)] which is nothing but P[(A1∩A2∩A3)] since (A1UA2UA3) contains (A1∩A2∩A3).

Now, you already have the value for P[(A1∩A2∩A3)], given and the denominator you already have calculated.

 

[Kromey]

Thank you. I was able to get the correct answer to be .01/.59 = .0169. However, I do not understand why it is the correct answer. Why does P(A1 ∩ A2 ∩ A3 | A1 ∪ A2 ∪ A3) = P(A1 ∩ A2 ∩ A3) / P(A1 ∪ A2 ∪ A3)?

Even if I see P(A | B) = P(A ∩ B) / P(B), I still don't fully understand.