# Binomial Problem!!

• May 31st 2010, 02:53 PM
seramika
Binomial Problem!!
Hello! I've been working on a lab this afternoon and this is my last question, I could really use the help! I'm not good at probability at all, so I'd like as much explanation as possible. Thank you! :)

A quality control engineer wants to check whether, in accordance with specifications, 90% of the products shipped are in perfect working condition. To this end, she randomly selects 12 items from each lot ready to be shipped and passes the lot only if all 12 are in perfect working condition. If one or more items are not in perfect working condition, she holds the lot for a complete inspection. Find the probabilities that she will commit the error of:

a) Holding a lot for complete inspection even though 90% of the items are in perfect working condition.
b) Letting a lot pass through even though only 80% of the items are in perfect working condition.
c) Letting a lot pass even though only 70% of the items are in perfect working condition.
• May 31st 2010, 06:24 PM
matheagle
Quote:

Originally Posted by seramika
Hello! I've been working on a lab this afternoon and this is my last question, I could really use the help! I'm not good at probability at all, so I'd like as much explanation as possible. Thank you! :)

A quality control engineer wants to check whether, in accordance with specifications, 90% of the products shipped are in perfect working condition. To this end, she randomly selects 12 items from each lot ready to be shipped and passes the lot only if all 12 are in perfect working condition. If one or more items are not in perfect working condition, she holds the lot for a complete inspection. Find the probabilities that she will commit the error of:

a) Holding a lot for complete inspection even though 90% of the items are in perfect working condition.
b) Letting a lot pass through even though only 80% of the items are in perfect working condition.
c) Letting a lot pass even though only 70% of the items are in perfect working condition.

(a) Here they found at least one defective out of the 12.

Let X be the number of defectives found, so $X\sim Bin(12, .1)$

$P(X\ge 1)=1-P(X=0)=1-(.1)^{12}$

(b) Here they found zero defectives out of the 12, so all 12 were good.

Let X be the number of nondefectives found, so $X\sim Bin(12, .8)$

$P(X=12)=(.8)^{12}$

(c) Again, they found zero defectives out of the 12, so all 12 were good.

Let X be the number of nondefectives found, so $X\sim Bin(12, .7)$

$P(X=12)=(.7)^{12}$
• May 31st 2010, 07:14 PM
seramika
Thank you so much!! :D