Question

there are 2 defective batteries in a group of 10. Five batteries are drawn at random (without replacement) from the group. What are the probabilities of getting 0 defective batteries in the sampel?

Answer 1

Probability is always what you want over total
So in this case what is the the number of good batteries at the beginning?
and the total number?

Answer 2

you should get 8/10
then for the second battery there are only 7 good ones left
and 9 left in total so 7/9

Answer 3

multiply this out and you got it
(8/10*7/9)*100% = percentage of getting two good batteries

Answer 4

for getting zero defective batteries in the sample: I thought the answer would be (2 nCr 0) times (8 nCr 2) divided by (10 nCr 5)

Answer 5

why? this not a combination problem

Answer 6

It's a probability, when doing combinations look for "Order doesn't matter" or other terms like that

Answer 7

Would the probability of getting 1 defective battery in the sample equal: (8/10) * (1/5)?

Answer 8

why 5? there are 10 at first
2 defective / 10 total
=2/10

Answer 9

you're right! the probability of defective batteries is 2/10. But if my sample is now 5 and I would like to choose only one that's defective doesn't that change the probability to 1/5?

Answer 10

how many are defective 1 or 2 that's the top number
how many in total, that's the bottom number

Answer 11

would the probability of getting 2 defective batteries in the sample be (2/10) * (1/10)?

Answer 12

or would it be 3/10?

Answer 13

both methods will work
\[\frac{\binom{8}{5}}{\binom{10}{5}}\] will get it

Answer 14

or
\[\frac{8\times 7\times 6\times 5\times 4}{10\times 9\times 8\times 7\times 6}\]

Answer 15

i think it is \(\frac{2}{9}\)

Answer 16

you can check that you get the same number either way

Answer 17

I got 2/9

Answer 18

for both methods

Answer 19

thanks satellite73