1. ## Probibility question

hi,

In a batch of 8,000 clock radios 10% are defective. A sample of 8 clock radios is randomly selected without replacement from the 8,000 and tested. The entire batch will be rejected if at least one of those tested is defective. What is the probability that the entire batch will be rejected? (Hint: First find the expected number of defective clock radios and then find P(At least 1 is defective).

help.

2. Outcomes are finite, 0 - 8. - Poisson is out of the question, though it may prove a useful approximation.

p = 0.10, the probability of failure, is nowhere near 0.50. Normal approximation probably isn;t very good.

A radio is either good or bad. - Maybe a Binomial model?

Okay, Binomial it is.

p(failure) = p = 0.10
p(working) = 1-p = 0.90 = q
n = 8

p(at least one defective) = 1 - p(none defective) = 1 - 0.90^8

Kind of a funny question, actually. If it is known to be a 10% defective shipment, why bother to test any? Either take it or don't. It's still 10%.

3. so the answer is 10% probability?

Originally Posted by TKHunny
Outcomes are finite, 0 - 8. - Poisson is out of the question, though it may prove a useful approximation.

p = 0.10, the probability of failure, is nowhere near 0.50. Normal approximation probably isn;t very good.

A radio is either good or bad. - Maybe a Binomial model?

Okay, Binomial it is.

p(failure) = p = 0.10
p(working) = 1-p = 0.90 = q
n = 8

p(at least one defective) = 1 - p(none defective) = 1 - 0.90^8

Kind of a funny question, actually. If it is known to be a 10% defective shipment, why bother to test any? Either take it or don't. It's still 10%.

4. ?? That's what I get for being chatty, I suppose.

Here is the most important part:

p(at least one defective) = 1 - p(none defective) = 1 - 0.90^8

Definitely not 10%

5. alternatively,

but is not as simple...

10% of 8000 is 800

that's the expected number of defective clocks

therefore, the expected number of non-defective clocks is 7,200

Hence, the probability of selecting 8 good clocks is

$\displaystyle \frac{\binom{7200}{8}}{\binom{8000}{8}}=\frac{7200 c_8}{8000c_8}=0.43$

therefore the probability of selecting at least one defective is

$\displaystyle 1-0.43$

6. Originally Posted by TKHunny
Outcomes are finite, 0 - 8. - Poisson is out of the question, though it may prove a useful approximation.

p = 0.10, the probability of failure, is nowhere near 0.50. Normal approximation probably isn;t very good.

A radio is either good or bad. - Maybe a Binomial model?

Okay, Binomial it is.

p(failure) = p = 0.10
p(working) = 1-p = 0.90 = q
n = 8

p(at least one defective) = 1 - p(none defective) = 1 - 0.90^8

Kind of a funny question, actually. If it is known to be a 10% defective shipment, why bother to test any? Either take it or don't. It's still 10%.
I'll just add that using the binomial distribution is an approximation to the Hypergeometric distribution.