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Q&A

Why aren't the "21 possibilities here" NOT equally likely?

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Please see the last sentence below, that I underlined in red.

Blitzstein. Introduction to Probability (2019 2 ed). Pp. 10-11.

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If the $36$ ordered pairs are equally likely, then the probability of getting a cake cone with chocolate in the afternoon and a waffle cone with vanilla in the evening is $P($cakeC,waffleV$)=1/36$, and the probability of getting a waffle cone with vanilla in the afternoon and a cake cone with chocolate in the evening is $P($waffleV,cakeC$)=1/36$.

If the $21$ possibilities (where the order of the cones is not important) are equally likely, then the probability of getting a cake cone with chocolate and a waffle cone with vanilla (in any order) is $1/21$.

But if the $36$ ordered pairs are equally likely, then the probability of getting a cake cone with chocolate and a waffle cone with vanilla (in any order) is just the sum of the two events $P($cakeC,waffleV$)+P($waffleV,cakeC$)$ (because the probability of two disjoint events is the sum of their probabilities). This is $2/36$. But $2/36\ne 1/21$, so the $21$ possibilities cannot be equally likely if the $36$ ordered pairs are equally likely.

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