Q.3.25 - A First Course in Probability

Question

 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

Step-by-Step Solution

Answer

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

Step1: The first radio that you check is defective

A - the radio is produced by A

B - the radio is produced by B

$D1$ the first radio is defective

$D_2$ - the second radio is defective

probabilities

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

D1 and D2 are independent Given A or B

Step2: The conditional probability that the other one is also defective?

Begin with the definition of conditional probabilities.

$P\left(D_2\mid D_1\right)=\frac{P\left(D_1{D}_2\right)}{P\left(D_1\right)}$

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

$P\left(D_2\mid D_1\right)=\frac{P\left(D_1{D}_2\mid A\right)P\left(A\right)+P\left(D_1{D}_2\mid B\right)P\left(B\right)}{P\left(D_1\mid A\right)P\left(A\right)+P\left(D_1\mid B\right)P\left(B\right)}$

Use conditions that are not dependent on the nominator.

$P\left(D_2\mid D_1\right)=\frac{P\left(D_1\mid A\right)P\left(D_2\mid A\right)P\left(A\right)+P\left(D_1\mid B\right)P\left(D_2\mid B\right)P\left(B\right)}{P\left(D_1\mid A\right)P\left(A\right)+P\left(D_1\mid B\right)P\left(B\right)}$

All of these probabilities are stated above, as follows:

$P\left(D_2\mid D_1\right)=\frac{{0.05}^2\cdot 0.5+{0.01}^2\cdot 0.5}{0.05\cdot 0.5+0.01\cdot 0.5}$

The calculation yields

$P\left(D_2\mid D_1\right)=\frac{13}{300}\approx 0.0433$