## anonymous one year ago I worked the probability problem I just need someone to check the answer please! In a batch of 50 batteries, four are defective. If we select 3 batteries, what is the probability that at least 2 are good.

1. anonymous

I did P=1-P(0 good)P(1good) my answer is 99%

2. amistre64

4 out of 50 are bad, so 46 out of 50 are good can we do a binomial approach?

3. amistre64

$\binom{50}{2}(\frac{46}{50})^2(\frac{4}{50})^{48}+\binom{50}{3}(\frac{46}{50})^{3}(\frac{4}{50})^{47}$

4. amistre64

i think 50s are 3 in the binomial code ...

5. anonymous

a binomial approach is used when looking for the probability of an exact number of things. i used the multiplication rule( that is what is used in the book to find at least prob.)

6. amistre64

$\binom{3}{2}(\frac{46}{50})^2(\frac{4}{50})^{1}+\binom{3}{3}(\frac{46}{50})^{3}(\frac{4}{50})^{0 }$

7. amistre64

exactly 2 or exactly 3 are at least 2

8. anonymous

P(at least 2 good)=1-69/4900*1/4900 =.999999 =99%

9. anonymous

P( 0 good) = ((4c3)(46c0))/(50c3) =1/4900

10. amistre64

it is a high number ... im getting 98.2 or so

11. anonymous

P(1 good)= ((4c2)(46c1))/(50c3) =69/4900

12. amistre64

shold we assume replacement or not? ggb = (4*46*47)/(50*49*48) gbg = (4*46*47)/(50*49*48) bgg = (4*46*47)/(50*49*48) ggg = (46*47*48)/(50*49*48)

13. anonymous

no replacement

14. amistre64

3*(4*46*47)/(50*49*48)+(46*45*44)/(50*49*48) without replacement gets us about 99.51

15. anonymous

So in other words 99% would be a correct answer

16. amistre64

im not the one grading it, but it seems fair enough :)

17. anonymous

Thanks!

18. amistre64

youre welcome