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

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

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}* 0.19^(0) * (1-0.19)^(7-0)+ \binom{7}{1}* 0.19^(1) * (1-0.19)^(7-1)`

=
`1 * 1 * 0.81^(7) +7 * 01.9^(1) * 0.81^(6)`

= 0.604

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