# Probability Help

• January 30th 2007, 05:33 PM
air2slb
Probability Help

:eek: 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????

• January 30th 2007, 07:27 PM
Soroban
Hello, air2slb!

Quote:

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

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

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

• January 31st 2007, 04:34 PM
air2slb
Im sorry I dont understand how you got the sample numbers. Could you show me that. Sorry!! Thanks!!:o
• February 1st 2007, 06:31 AM
AlvinCY
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