3. A lot of 15 parts contain 6 defective parts. Assume sampling without replacem

Question

3. A lot of 15 parts contain 6 defective parts. Assume sampling without replacement. (a) What is the probability that 1 defective part will be found in a sample of size 5? (b) What is the probability that atleast one defective part will be drawn in a sample of size of size 5? (c) Iffive parts are selected and you know the first four parts were defective, what is the probability that the 5th part drawn will also be defective? (d) A sample of size 5 is taken and each part drawn is recorded as being either good or defective and then replaced back in the lot and mixed before the be defective?

Explanation / Answer

a) one defective out of five

p(1 defective) * p(4 non defective)

6/15 * 9/14* 8/13 * 7/12* 6/11 [you can also write as 6C1/ 15C1* 9C4/14C4]

0.050

b) atleast 1 defective

1- p(no defective)

1 - ( 9/15 * 8/14 * 7/13 * 6/12 * 5/11)

1-0.0419

0.958

c) If the 4 parts were defective, it means out of 6 defective itmes 4 are already out. Hence the probability the the last one is also defective will be 2/11

d) This is a question of replacement.

p(atleast one defective)

1 - p(no defective)

1 - ( 9/15 * 9/15 * 9/15 * 9/15 * 9/15) {as they were put back in the lot, it is divided by 15 each}

1 - 0.07776

0.92224

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