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Numerade Educator

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Problem 69 Easy Difficulty

Show that $P(A \cap B | C)=P(A | B \cap C) \cdot P(B | C)$

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Video Transcript

All right, We're given this identity probability of a Intersect be given, See, because probability of a given be inter sexy times the probability of be given C Now, I'm gonna write out our multiplication rule to the site here and read clipping a little bit at the top. You'll have to forgive that I'm also going to write a different expression of it that gave us our definition of conditional probability. All right, now we're gonna work with this left side because it looks simpler using our definition of conditional probability. We can write it out like this. This equals the probability of a Intersect be in her sexy, all over the probability of C Because intersections associative, we can put the parentheses wherever we need them to. So we're gonna move them over now, On this numerator, we're gonna use our base form of our multiplication rule. So this becomes probability of oh, boy, a given probability are sorry. A given, be intersex. See times the probability a b intersect. See all over you. See, now I'm gonna rewrite this as the probability of a given be intersect. See, times probability Aby intersect. See all our probability of c. Wait a second. That looks like our definition of conditional probability. So we got probability of a given. Be intersect, See times probability of be given c. Well, what do you know? That's what we're trying to prove Q e d