# Probability question regarding independence

• Jun 26th 2012, 03:36 AM
usagi_killer
Probability question regarding independence
• Jun 26th 2012, 04:55 AM
HallsofIvy
Re: Probability question regarding independence
For (i) I would prefer to work with specific numbers. Suppose there are 100 boxes, each containing 100 screws (if the boxes do not all contain the same number of screws, this problem is impossible). 90% of the boxes have 1% defective screws- that is a total of 90 boxes each with 1 defective screw- a total of 90 defective screws. 9% of the boxes contain 10% defective screws- that is a total of 9 boxes each with 10 defective screws- at total of 90 defective screws. 1% of the boxes contain 100% defective screws- that is 1 box with 100 defective screws.

Altogether, then, there are 90+ 90+ 100= 280 defective screws out of (100)(100)= 10000 screws, so that 280/10000= 0.0280 or 2.8% are defective, as you say.
• Jun 26th 2012, 05:51 AM
biffboy
Re: Probability question regarding independence
You have started your 3rd line from the end with 0.01 when it should be 0.9
I agree with your first two answers but make the last answer 0.01099
For conditional probability I prefer to use the formula P(A/B) =P(intersection of A and B)/P(B)
• Jun 26th 2012, 08:01 AM
usagi_killer
Re: Probability question regarding independence
Thanks to both :)