Q. 3.4 - A First Course in Probability | StudyQuestionHub

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Q. 3.4

Question

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 $α_{i}$ . Show that the conditional probability that the ball is in box $j$ , given that a search of box $i$ did not uncover it, is

$\frac{P_{j}}{1 - α_{i} P_{i}} if j \neq i$

$\frac{(1 - α_{i}) P_{i}}{1 - α_{i} P_{i}} if j = i$

Step-by-Step Solution

Verified

Answer

$\frac{P_{j}}{1 - α_{i} P_{i}} if j \neq i$

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

Combining the above expressions we obtain the wanted probabilities:

$P (E_{j} ∣ A_{i}^{c}) = \frac{P_{j}}{1 - α_{i} P_{i}} for i \neq j$

$P (E_{i} ∣ A_{i}^{c}) = \frac{(1 - α_{i}) P_{i}}{P (A_{i}^{c})}$

1Step 1: Given Information

$E_{i}$ - the ball is in the $i$ -th box, $i \in {1, 2, \dots, n}$

$A_{i}$ - the ball is found in the $i$ -th box, $i \in {1, 2, \dots, n}$

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

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

$P (A_{i} ∣ E_{i}) = α_{i}$

$for i \in {1, 2, \dots, n}$

2Step 2: Explanation

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

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

$P (E_{j} ∣ A_{i}^{c})$

$P (E_{i} ∣ A_{i}^{c})$

Start with the definition

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

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

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

Formula for probability of a complement is

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

$A_{i} \subseteq E_{i}$ , and set operations show that:

$A_{i} \subseteq E_{i} \Rightarrow A_{i} = A_{i} E_{i}$

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

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

$P (A_{i}^{c}) = 1 - α_{i} P_{i}$

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

3Step 3: Final Answer

For $j \neq i$

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

$\to E_{j} \subseteq A_{i}^{c}, and$

$E_{j} \subseteq A_{i}^{c} \Rightarrow E_{j} A_{i}^{c} = E_{j}$

Therefore,

$2$

For $P (E_{i} ∣ A_{i}^{c})$

Use the identity

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

Transform the probabilities of intersection $P (E_{j} A_{i}^{c}) = P (E_{j})$ using conditional probability:

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

$\Rightarrow$

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

$\Rightarrow$

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

Combining the boxed expressions we obtain the wanted probabilities:

$P (E_{j} ∣ A_{i}^{c}) = \frac{P_{j}}{1 - α_{i} P_{i}} for i \neq j$

$P (E_{i} ∣ A_{i}^{c}) = \frac{(1 - α_{i}) P_{i}}{P (A_{i}^{c})}$

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