Batteries probability

[MGD]

A bag contains a total of 30 batteries, of which 6 are defective. Selecting 4 at random, without replacement, determine the probability that none of the batteries you select are good.

30 batt. = 6 defective

without replacement 30 x 29 x 28 x 27= 657,720

Lost, not sure where to go from here. How do you go about getting the probabilty?

 

[wjm11]

A bag contains a total of 30 batteries, of which 6 are defective. Selecting 4 at random, without replacement, determine the probability that none of the batteries you select are good.

30 batt. = 6 defective

without replacement 30 x 29 x 28 x 27= 657,720

Lost, not sure where to go from here. How do you go about getting the probabilty?

Chance of drawing a bad battery on the first draw: 6 out of 30
Second draw: 5 out of 29 (since there is no replacement, there are only 5 bad batteries left)
Third draw: 4 out of 28
Fourth draw: 3 out of 27

Multiply these together:

(6/30)(5/29)(4/28)(3/27) = ?

 

[soroban]

Hello, MGD!

Are you familiar with Combinations?

A bag contains a total of 30 batteries, of which 6 are defective.
Selecting 4 at random, without replacement, determine the
probability that none of the batteries you select are good.

\(\displaystyle \text{There are: }\:_{30}C_4 \:=\:{30\choose4} \:=\:\frac{30!}{4!\,26!} \:=\:27,\!405\text{ possible outcomes}\)

\(\displaystyle \text{We want 4 of the 6 defective batteries: }\;_6C_4 \:=\:{6\choose4} \:=\:\frac{6!}{4!\,2!} \:=\:15\text{ ways.}\)

\(\displaystyle \text{Therefore: }\(\text{4 defective}) \;=\;\frac{15}{27,\!405} \:=\:\frac{1}{1827}\)