Thread: Probability Help

Please help me with this. Im running short on time.


Sample of 4 calculators randomly selected from group contining 47 calculators that are defective and 25 with no defects. what is the probablility that all four calculators selected are defective????

Thank you in advance.

Hello, air2slb!

A sample of 4 calculators randomly selected from a group
containing 47 defective calculators and 25 that have no defects.
What is the probablility that all four calculators selected are defective?

There are: .$\displaystyle \binom{72}{4} = 1,028,790$ possible samples.

There are: .$\displaystyle \binom{47}{4} = 178,365$ samples with 4 defectives.


Therefore: .$\displaystyle P(\text{4 d{e}f}) \:=\:\frac{178,365}{1,027,790} \:=\:\frac{2585}{14,910}$

Im sorry I dont understand how you got the sample numbers. Could you show me that. Sorry!! Thanks!!

So you're selecting 4 calculators from a group of 72 calculators, without all the yucky bracket notation or combinations as some may know it

chance of the first being defective is 47/72,
chance of the second being defective is 46/71,
chance of the third being defective is 45/70, and the
chance of the fourth being defective is 44/69.

Multiply the four fractions and you should be able to get the same answer