Exercise 2.18, parts b and c. For the entirety of this problem, all cats are eit

Question

Exercise 2.18, parts b and c. For the entirety of this problem, all cats are either white or black, and all cats are adopted one-at-a-time from an animal shelter. Suppose a family has 2 cats, and assume that all combinations of black and white cats are equally likely, with the sample space describing the colors of each cat in the order that they were adopted by the family. Suppose we learn that there is a white cat in the family. (Precisely: we learn that there is at least one white cat.) What is the probability that the other cat is black? Suppose we see the family out with a cat, and the family members tell us that this it is their most-recently-adopted cat, and we see the cat is white. What is the probability that the older cat we have not yet seen is black? Now suppose there is a family with 4 cats, and all possibilities black and white cats are equally likely. We learn that the family has at least 3 white cat(s). What is the probability that the other 1 cat(s) are black? Now suppose we see the family from part (c) which has 4 cats out walking with 3 cat(s), all of whom are white, and the family members tell us that the cats we see are their 3 most-recently-adopted cats. What is the probability that the other 1 cat(s) that we have not yet seen are black?

Explanation / Answer

Let W shows the white Cat and B shows the black cat. So possible sample space for 2 cats is

S = {WW, WB, BW, BB}

(a)

P(at least one white cat) = 3/4

P(at least one white cat and one black cat) = 2/4

So the requried probaility is

P(one black cat|at least one white cat )= P(at least one white cat and one black cat) /P(at least one white cat) = 2/3

(b)

The probability that second is white, first can be white or black is

P(second is white) = 2/4

And

P(first is black and second is white) = 1/4

Therefore requried probabilty is

P(first is black | second is white) =P(first is black and second is white) / P(second is white) = 1/2 =0.5

(c)

Possible sample space for four cats:

S = {WWWW,WWWB,WWBW,WBWW,BWWW,WWBB,WBBW,BBWW,WBWB,BWWB,BWBW, WBBB,BWBB,BBWB,BBBW, BBBB}

P(at least 3 are white) = 5/16

P(at least 3 are white and one is black) = 4/16

So the required probability is

P(one is black|at least 3 are white)=P(at least 3 are white and one is black) /P(at least 3 are white) = 4/5 = 0.8

(d)

P(at least 3 are white) = 5/16

P(at least 3 are white and first is black) = 1/16

So the required probability is

P(first is black|at least 3 are white)=P(at least 3 are white and first is black) /P(at least 3 are white) = 1/5 = 0.2

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