Random probability: "Lisa bought a box of 40 AA batteries at a garage sale...."

[jmasta]

Random probability: "Lisa bought a box of 40 AA batteries at a garage sale...."

Lisa bought a box of 40 AA batteries at a garage sale. Her "loving" husband tested the batteries and tells her that 10 of them are bad (of course he mixed the bad batteries back in with the good). If Lisa were to randomly select 4 batteries for her calculator, what is the probability that all 4 are bad?

4/40 * 3/39 * 2/38 * 1/37

I feel like this isn't correct, can somehow direct me?

 

[Steven G]

Lisa bought a box of 40 AA batteries at a garage sale. Her "loving" husband tested the batteries and tells her that 10 of them are bad (of course he mixed the bad batteries back in with the good). If Lisa were to randomly select 4 batteries for her calculator, what is the probability that all 4 are bad?

4/40 * 3/39 * 2/38 * 1/37

I feel like this isn't correct, can somehow direct me?

There are 40 batteries--30 are good, 10 are bad. You want to pick 4 and have them all be bad. In how many ways can you do that? Please respond back with that answer that and if you can try to answer the question.

 

[HallsofIvy]

There are a total of 40 batteries and 10 of them are bad. What is the probability the very first battery selected, at random, is bad? If it is, then there are 39 batteries left and 9 of them are bad. What is the probability the second battery selected is bad? If it is, then there are 38 batteries left and 8 of them are bad. What is the probability the third battery selected is bad? If it is, then there are 37 batteries left and 7 of them are bad. What is the probability the fourth bttery selected is bad?

So what is the probabillty all four are bad?

 

[jmasta]

I think I see where I made the error.

So I reworked the problem and came to :

10/40 * 9/39 * 8/38 * 7/37 = 21/9139 or 0.00229784

So the probability would be 0.002

 

[Steven G]

I think I see where I made the error.

So I reworked the problem and came to :

10/40 * 9/39 * 8/38 * 7/37 = 21/9139 or 0.00229784

So the probability would be 0.002

Here is an alternate method to the way Prof Halls did the problem.

You can do it using the hypergeometric method.

You have two piles of batteries---10 bad and 30 good---- and you want to pick 4 and they all come from the bad pile.

Then the p(all 4 come from the bad pile is)= \(\displaystyle \dfrac{\dbinom{30}{0}\dbinom{10}{4}}{\dbinom{40}{4}}\)

The reason I prefer this method is because it works even if only some of the 4 batteries were bad unlike the other method.

 

[jmasta]

Thank you for the insight. Just to clarify, the way I did it the 2nd time was the correct answer though correct?

 

[Steven G]

I think I see where I made the error.

So I reworked the problem and came to :

10/40 * 9/39 * 8/38 * 7/37 = 21/9139 or 0.00229784

So the probability would be 0.002

10/40 * 9/39 * 8/38 * 7/37 is correct. As to whether it equals 0.00229784 I do not know (work it out yourself again, please).
I do know that if 10/40 * 9/39 * 8/38 * 7/37 = 0.00229784 then surely the probability will NOT equal 0.002. The probability will equal 0.00229784.