# Thread: sample are found defective the entire lot is rejected.

1. ## sample are found defective the entire lot is rejected.

A certain product is received in lots of 500 units. The quality control acceptance procedure is to choose 20 units at random from the lot and test them. If two or more units in the sample are found defective the entire lot is rejected.

What is the probability that a lot containing 40 defective units will be accepted? Can you see if I am on the right track Please help!

with n defectives by the total number of ways to choose any sample. There are
(40/n) ways to select the n defectrive units( 500/20-n) ways to select 920-n) good units and (500/20) ways to select any sample of 40 units
P(d =n) =(40/n) (500/20-n)/(500/20)

Since we want the probability that the sample will contain no more than two defective units, we simply sum
the probabilities of there being exactly zero, one, or two defectives in the sample, which gives us
P(d< 40) = P(d = 0) + P(d = 1) + P(d = 2) =

p(d<40) =P(d=0) + P(d=1)+ P(d=2)+ (400)(500/20) +(40/1) (500/3) =(40/2) =(500/2) all divided by 500/20

2. Originally Posted by masimini25
A certain product is received in lots of 500 units. The quality control acceptance procedure is to choose 20 units at random from the lot and test them. If two or more units in the sample are found defective the entire lot is rejected.

What is the probability that a lot containing 40 defective units will be accepted? Can you see if I am on the right track Please help!

with n defectives by the total number of ways to choose any sample. There are
(40/n) ways to select the n defectrive units( 500/20-n) ways to select 920-n) good units and (500/20) ways to select any sample of 40 units
P(d =n) =(40/n) (500/20-n)/(500/20)

Since we want the probability that the sample will contain no more than two defective units, we simply sum
the probabilities of there being exactly zero, one, or two defectives in the sample, which gives us
P(d< 40) = P(d = 0) + P(d = 1) + P(d = 2) =

p(d<40) =P(d=0) + P(d=1)+ P(d=2)+ (400)(500/20) +(40/1) (500/3) =(40/2) =(500/2) all divided by 500/20
You need to review the procedure for calculating selections in detail.
Also you haven't read the question clearly as you shouldn't be trying to count selections of 2 defectives.

The lot will be accepted if 20 good units are chosen from the 460 good ones (since 40 are defective)

or

if 19 good ones are selected from the 460 good and 1 defective is chosen from the 40 bad.

$\displaystyle \frac{\binom{460}{20}}{\binom{500}{20}}$

is the probability of choosing 20 good ones.

$\displaystyle \frac{\binom{460}{19}\binom{40}{1}}{\binom{500}{20 }}$

is the probability of choosing 19 good and 1 bad together

3. ## probability that a lot containing defective units

Thank you very much I appreciate it. I truly appreciate your fast response. I am still confused though what about the questions: What is the probability that a lot containing 40 defective units will be accepted?

4. The key is that "if 2 or more are found defective in the 20 tested, the entire lot is rejected".

This means the lot will not be rejected if zero defective are found or one defective are found. Or, if no more than one defective are found.
Hence you sum the probabilities of zero defective being discovered and one defective being discovered. That gives you the probability of the lot being accepted.

You don't add the probability of 2 defectives being discovered since the lot will be rejected in that case.