Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability help (1 Viewer)

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Could I just ask why in Q2, A and B are statistically independant, yet it in Q3, A and B, they aren't statistically independant?

I've been trying to rationalise that myself. Maybe deterministic is right.

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

So I guess the best way to calculate P(AB) in an exam, or anyother situation, would be look at any possible outcomes that match up with both Events A and B, i.e. outcomes that are the INTERSECTION of A and B, instead of using unreliable formulas such as P(AB) = P(A) * P(B)?

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Any confirmation advice please? thanks?

 

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

So I guess the best way to calculate P(AB) in an exam, or anyother situation, would be look at any possible outcomes that match up with both Events A and B, i.e. outcomes that are the INTERSECTION of A and B, instead of using unreliable formulas such as P(AB) = P(A) * P(B)?

There is nothing unreliable about the formula.

P(A & B) = P(A) x P(A|B)

When A and B are independent, P(A|B) simply becomes P(A) so that in this case P(A&B) = P(A)xP(B). As jm01 pointed out, conditional probability is not in the HSC syllabus. But intuitively, as in the case of the coloured-balls-in-the-urn example, we employ it without necessarily knowing that we have used it. You do not need to learn it but if you understand it, it will give you a better understanding.

 

[deterministic]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Only use P(AB)=P(A)*P(B) when you can actually tell A and B are physically independent events (for example toss of coin and roll of dice are physically independent, while drawing 2 cards without replacement from the same deck are not as the first draw affects the probability of second). Probability is not applying generic formulas to specific situations - you have to think about what is happening.

If you refer to my earlier examples, you are actually using conditional probability without knowing it (see the drawing of 2 specific cards from the same deck example)