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GE 331-Lecture 5

# GE 331-Lecture 5 - Total probability rule(cont For any...

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IE 300/GE 331 Lecture 5 Negar Kiyavash, UIUC 1 Total probability rule (cont) For any mutually exclusive events E 1 ,E 2 ,…,E n such that E 1 » E 2 » » E n =S ( exhaustive ) P(B) = P(B E 1 ) + … + P(B E n ) = P(B|E 1 )P(E 1 ) + … + P(B|E n )P(E n )

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IE 300/GE 331 Lecture 5 Negar Kiyavash, UIUC 2 Total probability rule (cont) Example: Box A has 3 green and 2 red balls, while Box B has 2 green and 2 red balls. A ball is drawn at random from Box A and transferred to Box B. Then, a ball is drawn at random from Box B. What is the probability that the ball drawn from Box B is green? A B
IE 300/GE 331 Lecture 5 Negar Kiyavash, UIUC 3 Total probability rule (cont) A B P(ball drawn from box B is green)=?

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IE 300/GE 331 Lecture 5 Negar Kiyavash, UIUC 4 Total probability rule (cont) Example: You and a friend are at a party with N–1 other people when suddenly a conga line forms. Assume that all (N+1)! orderings are possible. What is the probability that your friend is ahead of you in the line? Answer: 1/2 (by symmetry) Also we can show this by using total probability rule.
IE 300/GE 331 Lecture 5 Negar Kiyavash, UIUC 5 Example (cont) There are N+1 positions in the line All of which are equally likely with probability 1/(N+1) P(you are in j-th position) = 1/(N + 1) P(friend ahead|you in j-th) = (j – 1)/N Why j–1? Why N and not N+1? P(friend ahead) = P(friend ahead|you are at posn.1)P(you are at posn. 1) +…+P(friend ahead | you are at posn. N+1)P(you are at posn. N+1) 1/2 1)] (N N]/[N 1 [0 1)] [1/(N 1)/N] [(j ahead) P(friend 1 n 1 j = + + + + = + = + =

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