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Statistics / A First Course in Probability 9 / Chapter 3 / Problem 85P

A First Course in Probability | 9th Edition | ISBN: 9780321794772 | Authors: Sheldon Ross

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Repeat each of the 3 players selects from his own urn.

Chapter 3, Problem 85P

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QUESTION:

Repeat Problem 3.84 when each of the 3 players selects from his own urn. That is, suppose that there are 3 different urns of 12 balls with 4 white balls in each urn.

Problem 3.84.

An urn contains 12 balls, of which 4 are white. Three players—A, B, and C—successively draw from the urn, A first, then B, then C, then A, and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player if

(a) each ball is replaced after it is drawn;

(b) the balls that are withdrawn are not replaced.

Questions & Answers

QUESTION:

Repeat Problem 3.84 when each of the 3 players selects from his own urn. That is, suppose that there are 3 different urns of 12 balls with 4 white balls in each urn.

Problem 3.84.

An urn contains 12 balls, of which 4 are white. Three players—A, B, and C—successively draw from the urn, A first, then B, then C, then A, and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player if

(a) each ball is replaced after it is drawn;

(b) the balls that are withdrawn are not replaced.

ANSWER:

Step 1 of 2

We have 3 players A, B, C.

An urn contains 12 balls in that 12 are white.

a) The claim is to find the probability that each ball is replaced after it is drawn.

Probability of getting white ball is 4/12 = 1/3

Probability of A winning =Probability of A drawing white in his 1st , 2nd, 3rd draw………

= $\frac{1}{3}+\left[{\frac{2}{3}\times{}\frac{2}{3}\times{}\frac{2}{3}\times{}\frac{1}{3}}\right]+\left[{(\frac{2}{3}){\ }^{3}\times{}({\frac{2}{3}}_{\ }){\ }^{3}\times{}(\frac{1}{3})}\right]+..........$

= $\frac{1}{3}+\left[{(\frac{2}{3}){\ }^{3}\times{}\frac{1}{3}}\right]+\left[{(\frac{2}{3}){\ }^{6}\times{}(\frac{1}{3})}\right]+..........$

= $\frac{1/3}{1-(2/3){\ }^{3}}$

= $\frac{1/3}{19/27}$

= $\frac{9}{19}$

Then for three urns it is ( $\frac{9}{19}$ ${)}^{3}$

Hence Probability of A winning with replacement is 0.1062

Probability of B winning = $\frac{2}{3}\times{}\frac{1}{3}+[(\frac{2}{3}){\ }^{3}\times{}\frac{2}{3}\times{}\frac{1}{3}]+.......$

= $\frac{(2/3)(1/3)}{1-(2/3){\ }^{3}}$

= $\frac{6}{19}$

Then for three urns it is ( $\frac{6}{19}$ ${)}^{3}$

Hence the Probability of B winning with replacement is 0.03149.

The Probability of C winning =1-P(A winning)-P(B winning)

=1-( $\frac{9}{19}$ )-( $\frac{6}{19}$ )

= $\frac{4}{19}$

Then for three urns it is ( $\frac{4}{19}$ ${)}^{3}$

Hence Probability of C winning with replacement is 0.00933.

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