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(1 point) Consider another ubiquitous probability-course urn containing well-mix

ID: 3340707 • Letter: #

Question

(1 point) Consider another ubiquitous probability-course urn containing well-mixed black and white balls. There are 12 balls in total, 4 white and 8 black. 4 are chosen, one at a time and at random. Let Xi be 1 if the ith ball selected is white, and 0 otherwise. For parts (a) and (b), assume that the balls are selected without replacement. (a) Calculate the conditional probability mass function Xi given that X2 -1. PX1x, (111) = (b) Calculate the conditional probability mass function X1 given that X2 0 PXix, (010) = Px,x (410) For parts (c) and (d), assume that the balls are selected with replacement. (c) Calculate the conditional probability mass function X1 given that X2-1 Px, x (01) PX1x, (111) = (d) Calculate the conditional probability mass function Xi given that X20 px, lx, (010)- Px,x, (110)

Explanation / Answer

Solution :

Total number of balls in the Urn, n = 12

White balls = 4 and Black balls = 8

Also, Xi = 1 , if balls is White and 0 otherwise where i = selected ball

(a) P(X1 = 0 given that X2 = 1) = P ( X1 = 0 and X2 = 1)/P(X2 = 1) (without replacement)

P ( X1 = 0 and X2 = 1) means first ball is black and second white so probability = (8/12) * (4/11) = 8/33

P(X2 = 1) means first ball can be black or white so the required probability = (8/12)*(4/11) + (4/12)*(3/11) = 1/3

P(X1 = 0 given that X2 = 1) = (8/33)/(1/3) = 8/11

P(X1 = 1 given that X2 = 1) : P ( X1 = 1 and X2 = 1)/P(X2 = 1)

P(X1 = 1 given that X2 = 1)= ((4/12)*(3/11))/1/3 = 3/11

(b) Now, X2 = 0

P(X1 = 0 given that X2 = 0) = ((8/12)*(7/11))/((4/12)(8/11)+(8/12)*(7/11)) = 7/11

P(X1 = 1 given that X2 = 0) = ((4/12)*(8/11))/((4/12)(8/11)+(8/12)*(7/11)) = 4/11

(c) With replacement

P(X1 = 0 given that X2 = 1) : P( X1 = 0 and X2 = 1)/P(X2 = 1)

= ((8/12)(4/12))/((8/12)(4/12)+(4/12)(4/12)) = 2/3

P(X1 = 1 given that X2 = 1) : 1-2/3 = 1/3

(d) P(X1 = 0 given that X2 = 0) : P( X1 = 0 and X2 = 0)/P(X2 = 0)

= ((8/12)(8/12))/((8/12)(8/12)+(4/12)(8/12)) = 2/3

P(X1 = 1 given that X2 = 0) : 1-2/3 = 1/3

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