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An urn contains two white and two black balls. We draw balls from the urn randomly and stop after we find a black ball. What is the expectation of the total number of the drawn balls? ◦ $1\frac{1}{2}$$1\frac{2}{3}$$2$$2\frac{1}{3}$$2\frac{1}{2}$

My solution: Possible outcomes a. WWB b. WB c. B

$P(outcome a) = \frac{2}{4} * \frac{2}{3} * \frac{2}{2} = \frac{1}{3}$ $P(outcome b) = \frac{2}{4} * \frac{2}{3} * \frac{2}{2} = \frac{1}{3}$ $P(outcome c) = \frac{2}{4} = \frac{1}{2}$

Expectation = $\frac{1}{3} * \frac{1}{3} * \frac{1}{2} * 2 = 2\frac{1}{3}$

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  • $\begingroup$ Note that sum of outcomes probabilities cannot be $\frac13+\frac13+\frac12=\frac76>1$. $\endgroup$
    – NCh
    Jan 27, 2020 at 16:00

1 Answer 1

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probability we are done in first draw =$\frac{1}{2}$

probability we are done in 2nd draw=$\frac{1\times2}{2\times3}=\frac{1}{3}$

probability we are done at 3rd draw=$\frac{1}{6}$

Expectation =$\frac{1}{2}\times1+\frac{1}{3}\times2+\frac{1}{6}\times3$=$\frac{5}{3}$

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  • $\begingroup$ Why is probability that we're done at the third draw $\frac{1}{6}$? Isn't it WWB which is $\frac{2*2*2}{4*3*2}$? $\endgroup$ Jan 27, 2020 at 15:53
  • $\begingroup$ @RidwanSalahuddeen because that is the probability that the first draw was white, the second draw was also white given that the first was white, and the third draw is black given that the first two are white... which is calculated as $\frac{2}{4}\times\frac{1}{3}\times \frac{2}{2}=\frac{1}{6}$. Note that after the first white ball was drawn (it doesn't matter which), there will be one white ball out of three remaining balls at that point. $\endgroup$
    – JMoravitz
    Jan 27, 2020 at 15:54
  • $\begingroup$ first draw is white half probability half, 2nd is also white have probability one-third and now you are left with only black balls $\endgroup$ Jan 27, 2020 at 15:55

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