Chapter 10, Problem 48RE

Solution

(a)

To calculate: The probability that the lot of 10 light bulbs contains the no defective light bulbs.

The probability is 99.60%.

Given information:

Number of bulbs ( n )=10

Probability of success ( p )=0.0004

Formula used:

Pdf of the binomial distribution is:

  $P\left(X=k\right)=\left(\begin{array}{l}n\\ k\end{array}\right)\cdot p^k\cdot {\left(1-p\right)}^{n-k}$

Where,

k shows the number of successes.

Calculation:

The probability is calculated as:

  $\begin{array}{c}P\left(X=0\right)=\left(\begin{array}{l}10\\ 0\end{array}\right)\cdot {\left(0.0004\right)}^0\cdot {\left(1-0.0004\right)}^{10-0}\\ =\frac{10!}{0!\left(10-0\right)!}\cdot {\left(0.0004\right)}^0\cdot {\left(0.9996\right)}^{10}\\ =1\cdot 1\cdot {\left(0.9996\right)}^4\\ \approx 0.9960\\ =99.60\%\end{array}$

Thus, the probability that the lot of 10 light bulbs contains the no defective light bulbs is 99.60%.

(b)

To calculate: The probability that the lot of 10 light bulbs contains the two defective light bulbs.

The probability is 0.0007%.

Calculation:

The probability is calculated as:

  $\begin{array}{c}P\left(X=2\right)=\left(\begin{array}{l}10\\ 2\end{array}\right)\cdot {\left(0.0004\right)}^2\cdot {\left(1-0.0004\right)}^{10-2}\\ =\frac{10!}{2!\left(10-2\right)!}\cdot {\left(0.0004\right)}^2\cdot {\left(0.9996\right)}^8\\ =45\cdot {0.0004}^2\cdot {\left(0.9996\right)}^4\\ \approx 0.000007\\ =0.0007\%\end{array}$

Thus, the probability that the lot of 10 light bulbs contains the two defective light bulbs is 0.0007%.