Chapter 2.5, Problem 10E

(a)

To determine

Find the probability that the sample contains no defective fuses.

Answer to Problem 10E

The probability that the sample contains no defective fuses is 0.467.

Explanation of Solution

Calculation:

The conditional probability for any two events A, B with $P\left(A\right)>0$,

$P\left(B|A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$

Let $D_1$ denotes the first is defective, and $D_2$ denotes second is defective.

A batch of 10 fuses contains three defective ones.

The probability that the first one is defective is,

$P\left(D_1\right)=\frac{3}{10}$

The probability that the first one is not defective is,

$P\left(D_1^c\right)=\frac{7}{10}$

The probability that the second is defective given that first is defective is,

$P\left(D_2|D_1\right)=\frac{2}{9}$

The probability that the second is not defective given that first is not defective is,

$P\left(D_2^c|D_1^c\right)=\frac{6}{9}$

The probability that the second is not defective given that first is defective is,

$P\left(D_2^c|D_1\right)=\frac{7}{9}$

The probability that the sample contains no defective fuses is,

$\begin{array}{c}P\left(\text{no defective}\right)=P\left(D_2^c\cap D_1^c\right)\\ =P\left(D_2^c|D_1^c\right)P\left(D_2^c\right)\\ =\frac{6}{9}\times \frac{7}{10}\\ =0.467\end{array}$

Hence, the probability that the sample contains no defective fuses is 0.467.

(b)

To determine

Find the probability mass function of X.

Answer to Problem 10E

The probability mass function of X is,

X	0	1	2
Probability	0.467	0.466	0.067

Explanation of Solution

Calculation:

The formula for complement of event A is,

$P\left(A^c\right)=1-P\left(A\right)$

Let X denotes the random variable denoting the number of defective fuses in the sample.

The probability that the sample contains no defective fuses is,

$\begin{array}{c}P\left(\text{no defective}\right)=P\left(D_2^c\cap D_1^c\right)\\ =P\left(D_2^c|D_1^c\right)P\left(D_2^c\right)\\ =\frac{6}{9}\times \frac{7}{10}\\ =0.467\end{array}$

The probability that the sample contains defective fuses is,

$\begin{array}{c}P\left(X=2\right)=P\left(D_2\cap D_1\right)\\ =P\left(D_2|D_1\right)P\left(D_1\right)\\ =\frac{2}{9}\times \frac{3}{10}\\ =0.067\end{array}$

The probability that the sample contains one defective fuse is,

$\begin{array}{c}P\left(X=1\right)=1-P\left(X=0\right)-P\left(X=2\right)\\ =1-0.467-0.067\\ =0.466\end{array}$

Hence, the probability mass function of X is,

X	0	1	2
Probability	0.467	0.466	0.067

(c)

To determine

Find the probability that the defective fuse was the first one selected given that $X=1$.

Answer to Problem 10E

The probability that the defective fuse was the first one selected given that $X=1$ is 0.5.

Explanation of Solution

Calculation:

Bayes’ theorem:

The Bayes’ formula is,

$P\left(A_j|B\right)=\frac{P\left(A_j\right)P\left(B|A_j\right)}{{\displaystyle \sum _{i=1}^{k}P\left(A_i\right)P\left(B|A_i\right)}}$    for $j=1,2,\dots ,k$.

The probability that the defective fuse was the first one selected given that $X=1$ is,

$\begin{array}{c}P\left(D_1|X=1\right)=\frac{P\left(D_1\cap D_2^c\right)}{P\left(X=1\right)}\\ =\frac{P\left(D_2^c|D_1\right)P\left(D_1\right)}{P\left(X=1\right)}\\ =\frac{\left(\frac{7}{9}\times 0.3\right)}{0.466}\\ =\frac{0.233}{0.466}\end{array}$

                      $=0.50$

Hence, the probability that the defective fuse was the first one selected given that $X=1$ is 0.5.