# Probability with broken drainpipes

• Mar 3rd 2009, 09:41 AM
chella182
Probability with broken drainpipes
Sorry again for a vague-ish title - I have no clue where to begin with this.

To check the quality of plastic drainpipes a manufacturer selects 10 pipes at random from each large production batch and subjects them to an impact test. Flawed pipes break on the test but good ones do not. If more than two pipes on test break, the whole batch is scrapped. What are the probabilities:

a) that the batch will be scrapped when only 5% of the pipes in it are flawed;
b)that the batch will not be scrapped even if the proportion of flawed pipes is 25%?

There doesn't seem to be enough numbers here; I'm clearly missing something.
• Mar 3rd 2009, 10:27 AM
NoFace
Quote:

Originally Posted by chella182
Sorry again for a vague-ish title - I have no clue where to begin with this.

To check the quality of plastic drainpipes a manufacturer selects 10 pipes at random from each large production batch and subjects them to an impact test. Flawed pipes break on the test but good ones do not. If more than two pipes on test break, the whole batch is scrapped. What are the probabilities:

a) that the batch will be scrapped when only 5% of the pipes in it are flawed;
b)that the batch will not be scrapped even if the proportion of flawed pipes is 25%?

There doesn't seem to be enough numbers here; I'm clearly missing something.

For part a:

Seems to me like $X$ has a binomial distribution $BIN(10, 0.05)$ where 10 is the number of trials, and 0.05 is the success probability (probability of getting a defective one).
• Mar 3rd 2009, 10:46 AM
chella182
Ah thanks! Didn't think of it being binomial...

Maybe I'll have to approximate it using Poisson in this Q as well, since that's the only topic he's not yet covered in this assignment it would seem.

Thankyou :)
• Mar 3rd 2009, 03:00 PM
awkward
Quote:

Originally Posted by chella182
Sorry again for a vague-ish title - I have no clue where to begin with this.

To check the quality of plastic drainpipes a manufacturer selects 10 pipes at random from each large production batch and subjects them to an impact test. Flawed pipes break on the test but good ones do not. If more than two pipes on test break, the whole batch is scrapped. What are the probabilities:

a) that the batch will be scrapped when only 5% of the pipes in it are flawed;
b)that the batch will not be scrapped even if the proportion of flawed pipes is 25%?

There doesn't seem to be enough numbers here; I'm clearly missing something.

a) If 5% of the pipes in the batch of 10 are flawed, then the batch contains 2 flawed pipes. So exactly 2 pipes will break and the batch will not be scrapped.

b) Here you want P(X < 3) where X has a Binomial distribution with n = 10, p = 0.25.
• Mar 3rd 2009, 03:07 PM
sirellwood
haha, chella we r blatantley doing the same assignment. mas1301 yeah? have u done 2, 8(c), 10 or 11?
• Mar 4th 2009, 02:43 PM
chella182
Quote:

Originally Posted by awkward
a) If 5% of the pipes in the batch of 10 are flawed, then the batch contains 2 flawed pipes. So exactly 2 pipes will break and the batch will not be scrapped.

b) Here you want P(X < 3) where X has a Binomial distribution with n = 10, p = 0.25.

Thankyou for this :) really helpful.
• Mar 4th 2009, 06:00 PM
awkward
Quote:

Originally Posted by awkward
a) If 5% of the pipes in the batch of 10 are flawed, then the batch contains 2 flawed pipes. So exactly 2 pipes will break and the batch will not be scrapped.

b) Here you want P(X < 3) where X has a Binomial distribution with n = 10, p = 0.25.

a) I blew the simple arithmetic-- 5% of 10 is 0.5, not 2. (What could I have been thinking?) It is not possible to have 0.5 flawed pipes in the sample of 10. So I must not have understood the problem statement. I guess the author meant that 5% of the large production batch is flawed, not 5% of the sample. So the desired probability that the batch will be rejected is $P(X \geq 3)$ where X has a Binomial distribution with n = 10, p = 0.05.

b) Someone asked me off-line why P(X < 3) is the desired probability. It's because 3 or more broken pipes result in rejection of the batch. Note that $P(X < 3) = P(X \leq 2)$.
• Mar 5th 2009, 03:48 AM
chella182
Quote:

Originally Posted by awkward
a) I blew the simple arithmetic-- 5% of 10 is 0.5, not 2. (What could I have been thinking?) It is not possible to have 0.5 flawed pipes in the sample of 10. So I must not have understood the problem statement. I guess the author meant that 5% of the large production batch is flawed, not 5% of the sample. So the desired probability that the batch will be rejected is $P(X \geq 3)$ where X has a Binomial distribution with n = 10, p = 0.05.

b) Someone asked me off-line why P(X < 3) is the desired probability. It's because 3 or more broken pipes result in rejection of the batch. Note that $P(X < 3) = P(X \leq 2)$.

Thanks :) now this is helpful (Rofl) only joking (Wink)