Chapter 3: Q.3.25 (page 111)

Two local factories, A and B, produce radios. Each radio produced at factory A is defective with probability $. 05$ whereas each one produced at factory B is defective with probability $. 01$ . Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory A or factory B. If the first radio that you check is defective, what is the conditional probability that the other one is also defective

Short Answer

Expert verified

The factory from which the radios were sourced. The defectiveness of the two radios is unrelated to the factory from which they were produced. $\frac{13}{300} \approx 0.0433$

Step by step solution

The first radio that you check is defective

A - the radio is produced by A

B - the radio is produced by B

$D 1$ the first radio is defective

$D_{2}$ - the second radio is defective

probabilities

$\begin{matrix} P (A) = 0.5 \\ B = A^{c} \end{matrix}$
$i = 1, 2$

D1 and D2 are independent Given A or B

The conditional probability that the other one is also defective?

Begin with the definition of conditional probabilities.

$P (D_{2} ∣ D_{1}) = \frac{P (D_{1} D_{2})}{P (D_{1})}$

Condition both nominator and the denominator on A or B (since $B = A^{c}$ ):

$P (D_{2} ∣ D_{1}) = \frac{P (D_{1} D_{2} ∣ A) P (A) + P (D_{1} D_{2} ∣ B) P (B)}{P (D_{1} ∣ A) P (A) + P (D_{1} ∣ B) P (B)}$

Use conditions that are not dependent on the nominator.

$P (D_{2} ∣ D_{1}) = \frac{P (D_{1} ∣ A) P (D_{2} ∣ A) P (A) + P (D_{1} ∣ B) P (D_{2} ∣ B) P (B)}{P (D_{1} ∣ A) P (A) + P (D_{1} ∣ B) P (B)}$

All of these probabilities are stated above, as follows:

$P (D_{2} ∣ D_{1}) = \frac{{0.05}^{2} \cdot 0.5 + {0.01}^{2} \cdot 0.5}{0.05 \cdot 0.5 + 0.01 \cdot 0.5}$

The calculation yields

$P (D_{2} ∣ D_{1}) = \frac{13}{300} \approx 0.0433$

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The first radio that you check is defective

The conditional probability that the other one is also defective?

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