1) Consider an urn containing 4 black balls and 8 white balls, and consider rand

Question

1) Consider an urn containing 4 black balls and 8 white balls, and consider randomly and independently
drawing, with replacement, a ball from this urn 9 times.

(a) What is the probability of selecting a black ball exactly two times (in 9 attempts)? (Round the probability
to the nearest thousandth.) (Note: Making the draws with replacement results in the probability
of obtaining a black ball be the same value on every draw, regardless of the outcomes of any other
draws.)

(b) What is the probability of selecting a black ball at least two times (in 9 attempts)? (Round the
probability to the nearest thousandth.)

(c) What is the expected number of times a black ball will be drawn?

Explanation / Answer

4 of 12 balls are black, so P(black) = 1/3

Then, as we select with replacement 9 times,

P(X blacks) = C(9,x)(1/3)^x(2/3)^(9-x)

Thus, the table below shows the probabilities:

a)P(2 blacks) = C(9,2)*(1/3)^2(2/3)^(9-2)=36*0.111111*0.058528=0.234

b)P(at least 2) = 1 - P(less than 2) = 1 - P(0) - P(1) = 1 -0.026012 -0.117055 =

.856932 = .857

c) np = 9(1/3) = 3

X C(9,X) (1/3)^X (2/3)^(9-x) Prob. 0 1 1 0.026012 0.026012 1 9 0.333333 0.039018 0.117055 2 36 0.111111 0.058528 0.234111 3 84 0.037037 0.087791 0.273129 4 126 0.012346 0.131687 0.204847 5 126 0.004115 0.197531 0.102423 6 84 0.001372 0.296296 0.034141 7 36 0.000457 0.444444 0.007316 8 9 0.000152 0.666667 0.000914 9 1 5.08E-05 1 5.08E-05 Total 1

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