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Unformatted text preview: Math 170
Kouba
The Law of Total Probability B e
RECALL 1 A
I.) An(BUC) = (AnB)U(AnC')
I1-) P(BIC) = 512% —» P(B n C) = P(B|C) - 19(0) O c.
111.) If B n C = as, then. P(B U C) = 19(3) + 19(0) . B Q Let Q be a sample space and let B1, 32, B3, - - - , 3,, be a partition for 0. Let A be an event
in 9. Then 33% A = A09
= A 0 (31 U 32 U 33 U - - - U B") (since 81,32,33, - - - , Bn are a partition for Q)
= (AflBl)U(AflB2)U(AflBg)U---U(AflBn) (by RECALL I.) ——+
P(A) = P(A n Bl) + P(A n 32) + - - . + P(A n Bn) (by RECALL III.)
= P(A|B1) - P(Bl) + P(A|Bg) - P(Bz) + - - - + P(A|Bn) - P(Bn) (by RECALL 11.), Le, P(A) = Z?=1P(A n Bi) = 2?=1P(A|Bi) - P(Bi) . This is called the Law of Total Probability. ...
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- Summer '09
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