Question

3% of a given lot of manufactured parts are defective,wht is the probability that in a sample of 4 items none will be defective.​

Answer 1

Probability that in a sample of 4 items none will be defective is 0.8853.

Step-by-step explanation:

We are given that 3% of a given lot of manufactured parts are defective.

A sample of 4 items is chosen at random.

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 = 4 items

            r = number of success = none is defective

           p = probability of success which in our question is % of a given  

                 lot of manufactured parts that are defective, i.e; 3%

LET X = Number of items that are defective

So, it means X ~ $Binom(n=4, p=0.03)$

(i) Probability that in a sample of 4 items none will be defective is given by = P(X = 0)

                P(X = 0)  =  $\binom{4}{0}\times 0.03^{0} \times (1-0.03)^{4-0}$

                               =  $1 \times 1 \times 0.97^{4}$

                               = 0.8853

Therefore, the required probability is 0.8853.