Question

At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time. A random sample of 7 parts produced by this machine is chosen. Find the probability that fewer than 2 of these parts are defective. Do not round your intermediate computations, and round your answer to three decimal places

Answer 1

Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 7 parts

            r = number of success = fewer than 2

           p = probability of success which in our question is % of defective

                 parts produced by one of the machine, i.e; 19%

LET X = Number of parts that are defective

So, it means X ~ Binom(n = 7, p = 0.19)

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

    P(X < 2) = P(X = 0) + P(X = 1)

                  =  $\binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}$

                  =  $1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}$

                  =  0.604

Therefore, the probability that fewer than 2 of these parts are defective is 0.604.