Question

Electronic baseball games manufactured by Tempco Electronics are shipped in lots of 36. Before shipping, a quality-control inspector randomly selects a sample of 7 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 2 defective games will still be shipped? (Round your answer to three decimal places.)

Answer 1

hypergeometric distribution

Answer 2

Proportion defective = 2/36 = p
P is the chance of acceptance = P
\[P=(1-p)^{7}=(1-\frac{2}{36})^{7}=0.67\]

Answer 3

I assume the sampling is done without replacement. ? ?

Answer 4

This is single sampling.

Answer 5

is the sampling done where you pick 7 different items...or is it possible to select the same item more than once.

Answer 6

In this type of sampling the sample is selected randomly (say using a random number generator). If the same number comes again during the selection process then discard the result and select again

Answer 7

then the sampling is done without replacement...use the hypergeometric distribution

Answer 8

I understand your reply based on my answer to your question about the method of sampling. My answer is wrong.
To ensure that sampling is truly random a table of random numbers should be used. These consist of digits in an order which has been tested and found to be free from any systematic defects. A number is assigned to each unit liable to be tested. Then reference to the table will ensure a random selection is made. The question states that that the inspector "randomly selects a sample".
If truly random sampling was used then I believe my answer to the question is correct.

Answer 9

your answer is correct if we are randomly sampling with replacement

Answer 10

The question deals with lot acceptance sampling without replacement. The lot size is small so in this case you are right, the hypergeometric distribution applies.

Answer 11

I used the hypergeometric calculator at Stat Trek. This calculator gave the hypergeometric probability of finding one defective game as follows:
P(x=1) = 0.3222
Therefore the probability that the defective lot will be shipped is:
P(acceptance) = 1 - 0.3222 = 0.678

Answer 12

the probability is 29/45

Answer 13

Zarkon, Did you calculate using matrices?

Answer 14

\[\frac{{34\choose 7}{2\choose 0}}{{36\choose 7}}\]

Answer 15

\[\frac{\left(\begin{matrix}34 \\ 6\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right)}{\left(\begin{matrix}36\\ 7\end{matrix}\right)}\]
I expected this.

Answer 16

the question is ...What is the probability that a lot containing exactly 2 defective games will still be shipped?

in order to be shipped there can't be any defectives...therefore out of the 2 defective we want 0 of them