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Q.4.10

Expert-verified

A First Course in Probability

Found in: Page 163

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

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Short Answer

Let $X$ be the winnings of a gambler. Let $p (i) =$ $P (X = i)$ and suppose that
$p (0) = 1 / 3; p (1) = p (- 1) = 13 / 55$
$p (2) = p (- 2) = 1 / 11; p (3) = p (- 3) = 1 / 165$
Compute the conditional probability that the gambler wins $i, i = 1, 2, 3,$ given that he wins a positive amount.

The probabilities are:

$ℙ (X = 1 ∣ Y) = \frac{39}{55}, ℙ (X = 2 ∣ Y) = \frac{3}{11}, ℙ (X = 3 ∣ Y) = \frac{1}{55}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Given information

Let $X$ be the winnings of a gambler. Let $p (i) = P (X = i)$ and suppose that

p(0) = 1/3; p(1) = p(−1) = 13/55;

p(2) = p(−2) = 1/11; p(3) = p(−3) = 1/165

Step 2:Explanation

Let $X$ be a winning of the gambler. Also let us define $p (i) = ℙ (X = i)$ and suppose that $p (0) = \frac{1}{3}, p (1) =$ $p (- 1) = \frac{13}{55}, p (2) = p (- 2) = \frac{1}{11}, p (3) = p (- 3) = \frac{1}{165}$ We are to calculate conditional probability of gambler winnning $i = 1, 2, 3$ given that he wins positive amount.

Firstly let us calculate the probability that he won a positive amount.

$ℙ (Y) = p (1) + p (2) + p (3) = \frac{13}{55} + \frac{1}{11} + \frac{1}{165} = \frac{55}{165} = \frac{1}{3}$

Therefore we have:

$ℙ (X = 1 ∣ Y) = \frac{ℙ (X = 1, Y)}{ℙ (Y)} = \frac{ℙ (X = 1)}{ℙ (Y)} = \frac{\frac{13}{55}}{\frac{1}{3}} = \frac{39}{55}$

$ℙ (X = 2 ∣ Y) = \frac{ℙ (X = 2, Y)}{ℙ (Y)} = \frac{ℙ (X = 2)}{ℙ (Y)} = \frac{\frac{1}{11}}{\frac{1}{3}} = \frac{3}{11}$

$ℙ (X = 3 ∣ Y) = \frac{ℙ (X = 3, Y)}{ℙ (Y)} = \frac{ℙ (X = 3)}{ℙ (Y)} = \frac{\frac{1}{165}}{\frac{1}{3}} = \frac{1}{55}$

Therefore, we are done.

Step 3:Final answer

The probabilities are

$ℙ (X = 1 ∣ Y) = \frac{39}{55}, ℙ (X = 2 ∣ Y) = \frac{3}{11}, ℙ (X = 3 ∣ Y) = \frac{1}{55}$

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