Question

A batch of 400 sewing machines contains 6 that are defective.

(a) Three sewing machines are selected at random, without replacement. What is the probability that the second sewing machine is defective, but the first and third are acceptable?

(b) Three sewing machines are selected at random, without replacement. What is the probability that the third sewing machine is defective, given that the first two are acceptable?

(c) Sewing machines are selected, this time with replacement. What is the probability that the tenth machine selected is the first defective one?

(d) Sewing machines are selected, with replacement. What is the expected number of machines selected until you find two defective ones?

Answer 1

In our problem, we have 6 defective machines among 400.

so, the probability of accepted machine=q=P(accepted)=(400-6)/400=0.985

and p=P(defective)=6/400=0.015

(a) Three sewing machines are selected at random, without replacement. What is the probability that the second sewing machine is defective, but the first and third are acceptable?

Here to find the probability of,

P( accepted, defective, accepted)

=0.01462595

(b) Three sewing machines are selected at random, without replacement. What is the probability that the third sewing machine is defective, given that the first two are acceptable?

Here to find the probability of,

P( accepted, accepted, defective)

= 0.01462595

(c) Sewing machines are selected, this time with replacement. What is the probability that the tenth machine selected is the first defective one?

Now we need to find accepted machines are first 9 and 10th in defective.

To do this we define a random variable X, denotes number of drawings required to get first defective.

so, PMF fo X is

P(X=x)=p qx-1=(0.015) 0.985x-1

, x=1,2,3,...........,This is called Geometric distribution

Here our task is to find,

P(X=10)=0.01309234

(d) Sewing machines are selected, with replacement. What is the expected number of machines selected until you find two defective ones?

So, we take two samples from the above geometric distribution X and Y

to find E(X)

We require

E(X+Y)=2/p=133.3333

(answer)