Question

A production lot of 80 units has 8 defective items. We draw a random sample of 10 units and we want to know:
a.- the probability that the sample contains less than 3 defective articles
b.- the probability that the sample contains at least 3 good articles
c.- the probability that the sample contains more than 6 good articles

Answer 1

Let p be the success of getting a defective item.

Since , we are having 8 out of 80 defective.

Thus , Probability of an item being defective is (8/80) or 0.1.

Now , it is a binomial distribution with n = 10 , p = 0.1

Using Binomial Distribution,

P( x = r) = p^r (1 - p)^n-r

a) P( x < 3) = P( x =0) + P( x = 1) + P(x = 2)

= (0.1)⁰(1 - 0.1)^10-0 + (0.1)¹(1-0.1)^10-1 + (0.1)² (1-0.1)^{10 - 2}

= (0.9)¹⁰ + 10*(0.1)*(0.9)⁹ + 45*(0.1)² (0.9)⁸

= 0.3487 + 0.3874 + 0.1937

= 0.9298

Thus , there is 0.9298 probability that less than 3 out of 10 will be defective items.

b) Probability that atleast 3 will be non-defective products.

In other words, we need to find probability of getting less than 3 defective item .

P( x < 3) = P( x=0) + P( x =1) + P(x =2)

Which is same as part 1 .

Thus, there is 0.9298 probability that there will be atleast 3 defective items.

c) Probability of getting more than 6 good articles.

Thus , we need to find Probability of getting less than or equal to 6 defective articles.

P( x 6) = P(x=0) + P(x=1)...+ P(x=6)

Using Appendix Tables, A.1 :

P (x 6) = 1.000 (approx)

Thus , there is certainty that there will be more than 6 good articles.

Please like the answer,Thanks!