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Numerade Educator

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Problem 63 Medium Difficulty

Show that for any three events $A, B,$ and $C$ with $P(C)>0, P(A \cup B | C)=P(A | C)+P(B | C)-P(B | C)-P(B | C)-P(B | C)-P(B | C)-P(B | C)-P(B | C)-$ $P(A \cap B | C) .$

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Video Transcript

all right, we're given this identity to prove the probability of a union Be given, See is he goes to the probability of a given C plus a probability of be given C minus the probability of a intersects beep Given c for this, we're gonna use to hard probability identities, one of which is the multiplication I didn t which is as follows the probability of a intersect be It was the probability of a given b times the probability of b what we're going to do. So we're going to rewrite this probability a given B Because the probability a intersect to be over probability of B the second rule will need is our additional probability of a union be It was the probability of a it's probability of B minus the probability of a intersects B All right, so let's use his modified multiplication rule on the left hand side with this identity probability of a union be given. See those probability of a union, be intersex, see all over the probability of C intersection quote unquote distributes. So this because probability of a intersex see union be in our sexy over probability. See, now we're going to use our addition rule to expand this out of little bits of probability of And her sexy lost the probability of be her sexy minus the probability of, well, normally it be a and her sexy intersect be in her sexy. But that just is a intersects b intersect. See, we don't need to be redundant to repeat ourselves. We're gonna separate this into three separate fractions using the same denominator. All right as follows. Sorry. That should be a minus. Now, we're going to use our modified multiplication rule once again. So this becomes probability of a given c. This becomes probability of being her sexy. And because intersection is associative, we can put parentheses wherever he wants. Are gonna put prince sees around, are a intersect, Be to make this intersect, to be given C, and thus we've gotten to the right side of our identity. Q e d