Chapter 10, Problem 45RE

Solution

(a)

To calculate: The probability that the lot of four bats contains the no defective bats.

The probability is 92.24%.

Given information:

Four bats are selected at random.

Number of trials( n )=4

Probability of producing defective baseball bat $\left(p\right)=0.02$

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 computed as:

  $\begin{array}{c}P\left(X=0\right)=\left(\begin{array}{l}4\\ 0\end{array}\right)\cdot {\left(0.02\right)}^0\cdot {\left(1-0.02\right)}^{4-0}\\ =\frac{4!}{0!\left(4-0\right)!}\cdot {\left(0.02\right)}^0\cdot {\left(0.98\right)}^4\\ =1\cdot 1\cdot {\left(0.98\right)}^4\\ \approx 0.9224\\ =92.24\%\end{array}$

Thus , the probability that the lot of four bats contains the no defective bats is 92.24%.

(b)

To calculate: The probability that the lot of four bats contains the one defective bats.

The probability is 7.53%.

Calculation:

The probability is calculated as:

  $\begin{array}{c}P\left(X=1\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)\cdot {\left(0.02\right)}^1\cdot {\left(1-0.02\right)}^{4-1}\\ =\frac{4!}{1!\left(4-1\right)!}\cdot {\left(0.02\right)}^1\cdot {\left(0.98\right)}^3\\ =4\cdot (0.02)\cdot {\left(0.98\right)}^3\\ \approx 0.0753\\ =7.53\%\end{array}$

Thus, the probability that the lot of four bats contains the one defective bat is 7.53%.