Probability Assumption

[keep_running]

Hello,
Question on the theory of probability:
If A and B are mutually exclusive, then shouldn’t P(AB) still equal P(A) * P(B)?
My reasoning is that P(AB) = 0 based on the fact that this would not exist if A and B are mutually exclusive. However, wouldn’t P(A) * P(B) also be 0, since this does not exist as well? That would make P(AB) = P(A) * P(B)..
In the book, it says they are not equal.
Thank you!

 

[S2000magician]

keep_running wrote:If A and B are mutually exclusive, then shouldn’t P(AB) still equal P(A) * P(B)?

No.
Consider a fair coin: P(H) = P(T) = 0.5.
The probability of getting both heads and tails on the same toss is: P(HT) = 0.
However, P(H) × P(T) = 0.5 × 0.5 = 0.25 ≠ 0.
Candidates often confuse independence with mutual exclusivity. If A and B are independent, knowing that A happens tells you nothing about whether B happens. If A and B are mutually exclusive, knowing that A happens tells you that B did not happen.

 

[keep_running]

From your scenario, if you got P(H), couldn’t this lead you to assume that P(T) = 0 (since they are mutually exclusive)?
Then, P(H) x P(T) = 0 = P(HT)

 

[S2000magician]

keep_running wrote:From your scenario, if you got P(H), couldn’t this lead you to assume that P(T) = 0 (since they are mutually exclusive)?
Then, P(H) x P(T) = 0 = P(HT)

I’m not sure what you mean by “if you got P(H)”.
Perhaps you mean “if you toss a heads”.
If so, then you’re confusing the ex ante probabilities (before you toss the coin) with the ex post probabilities (after you toss the coin). The analysis should all be ex ante.
Before we toss the coin, P(H) = P(T) = 0.5, and P(HT) = 0.