I know that there is the addition rule of probability,

Esther Hoffman 2022-04-30 Answered
I know that there is the addition rule of probability, but I want to understand the intuition behind it. Specifically, why does OR signifies addition in probability theory?
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Answers (2)

Brianna Sims
Answered 2022-05-01 Author has 19 answers
I suppose that you question is this:

If two events A or B cannot occur simultaneously, then why is the probability that A occurs or B occurs equal to the probability that A occurs plus the probability that B occurs?

Suppose that there are, say, 100 possibilities, that A takes place in 50 of them and that B takes place in 20 of them. Then the probability that A occurs is 1 2 ( = 50 100 ) and the probability that B occurs is 1 5 ( = 20 100 ) . What is the probability that A occurs or B occurs? Well, out of those 100 possibilities, A occurs or B occurs exactly in 70 of them (this is where I use the fact that A or B cannot occur simultaneously). So, the probability that A occurs or B occurs is
70 100 = 50 100 + 20 100 = probability that  A  occurs + probability that  B  occurs.
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Tyler Velasquez
Answered 2022-05-02 Author has 19 answers
Because the probability is the number of favorable draws over the total number of draws.
And the number of favorable draws are additive: the number of [red or green] balls is the number of red plus the number of green.

The additive rule is only valid for disjoint categories. For example, if you have black/white balls and dice, it is not necessarily true that
#(black or ball) = #black + #balls .
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What is P ( A ( B C ) )?

The question says it all. I know
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Would this mean,
P ( A ( B C ) ) = P ( A B ) + P ( A C ) ?
Just want to make sure.
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11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6
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20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?

I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:

1) the first two are yellow
2) the first two are blue

I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]