Q. 3.4 A ball is in any one of n聽boxes... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 3: Q. 3.4 (page 106)

A ball is in any one of $n$ boxes and is in the $i$ th box with probability $P_{i}$ . If the ball is in box $i$ , a search of that box will uncover it with probability $伪_{i}$ . Show that the conditional probability that the ball is in box $j$ , given that a search of box $i$ did not uncover it, is
$\frac{P_{j}}{1 - 伪_{i} P_{i}} if j 鈮� i$
$\frac{(1 - 伪_{i}) P_{i}}{1 - 伪_{i} P_{i}} if j = i$

Short Answer

Expert verified

$P (E_{j} A_{i}^{c}) = P (E_{j})$

$P (E_{i} A_{i}^{c}) = P_{i} (1 - 伪_{i})$

Combining the above expressions we obtain the wanted probabilities:

$P (E_{j} 鈭� A_{i}^{c}) = \frac{P_{j}}{1 - 伪_{i} P_{i}} for i 鈮� j$

$P (E_{i} 鈭� A_{i}^{c}) = \frac{(1 - 伪_{i}) P_{i}}{P (A_{i}^{c})}$

Step by step solution

Given Information

$E_{i}$ - the ball is in the $i$ -th box, $i 鈭� {1, 2, 鈥�, n}$

$A_{i}$ - the ball is found in the $i$ -th box, $i 鈭� {1, 2, 鈥�, n}$

Probabilities (if the $i$ -th box is searched):

$P (E_{i}) = P_{i}$

$P (A_{i} 鈭� E_{i}) = 伪_{i}$

$for i 鈭� {1, 2, 鈥�, n}$

Explanation

Also, $A_{i} 鈯� E_{i}$ , that is, the ball can only be found in the $i$ -th box if it is there.

And $E_{i} 鈭� E_{j} = 鈭�$ for $i 鈮� j$ , they are mutually exclusive because the ball can only be in one box. In these terms, the stated probabilities correspond to:

$P (E_{j} 鈭� A_{i}^{c})$

$P (E_{i} 鈭� A_{i}^{c})$

Start with the definition

$P (E_{j} 鈭� A_{i}^{c}) = \frac{P (E_{j} A_{i}^{c})}{P (A_{i}^{c})}$

$P (E_{i} 鈭� A_{i}^{c}) = \frac{P (E_{i} A_{i}^{c})}{P (A_{i}^{c})}$

$P (A_{i}^{c})$

Formula for probability of a complement is

$P (A_{i}^{c}) = 1 - P (A_{i})$

$A_{i} 鈯� E_{i}$ , and set operations show that:

$A_{i} 鈯� E_{i} 鈬� A_{i} = A_{i} E_{i}$

This renders the former formula

$P (A_{i}^{c}) = 1 - P (A_{i} E_{i})$

Transform this probability of intersection using conditional probability to obtain

$P (A_{i}^{c}) = 1 - P (A_{i} 鈭� E_{i}) P (E_{i})$

$P (A_{i}^{c}) = 1 - 伪_{i} P_{i}$

This is the denominator of fractions ( $1$ ) and ( $2$ )

Final Answer

For $j 鈮� i$

If the ball was in $j$ -th box, it could not have been found in $i$ -th box

$鈫� E_{j} 鈯� A_{i}^{c}, and$

$E_{j} 鈯� A_{i}^{c} 鈬� E_{j} A_{i}^{c} = E_{j}$

Therefore,

$P (E_{j} A_{i}^{c}) = P (E_{j})$

For $P (E_{i} 鈭� A_{i}^{c})$

Use the identity

$P (E_{i}) = P (E_{i} A_{i}) 鈭� P (E_{i} A_{i}^{c})$

Transform the probabilities of intersection $E_{i} A_{i}$ using conditional probability:

$P (E_{i}) = P (A_{i} 鈭� E_{i}) P (E_{i}) + P (E_{i} A_{i}^{c})$

$鈬�$

$P_{i} = 伪_{i} P_{i} + P (E_{i} A_{i}^{c})$

$鈬�$

$P (E_{i} A_{i}^{c}) = P_{i} (1 - 伪_{i})$

Combining the boxed expressions we obtain the wanted probabilities:

$P (E_{j} 鈭� A_{i}^{c}) = \frac{P_{j}}{1 - 伪_{i} P_{i}} for i 鈮� j$

$P (E_{i} 鈭� A_{i}^{c}) = \frac{(1 - 伪_{i}) P_{i}}{P (A_{i}^{c})}$

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