Question

A store has 80 parts in its inventory. 30 are from source A, 50 from source B. 20% of A are defective, 8% of B are defective. Whats the probability that exactly 2 from a selection of 5 w/out replacement are defective?

Answer 1

30(.2) + 50(.08)

6 + 4, 10 defective total

P(defect) = 10/80

5C2 = 10, sooo

10(10*9*(78)*(77)*(76))/(80*79*78*77*76)

10(90)/(80*79) whats wrong with tis approach?

Answer 2

.1424 isnt an option on the practice exam ...

Answer 3

i dont think it matters where the parts come from after they are in inventory, so picking a part thats defective is 10/80 and 9/79 regardless

Answer 4

i have 80

30(.2) are bad, and 50(.08) are bad

the sources defined the number that are bad from each one.

i already know how many are bad regardless of the source since I have all the parts in my possession and i am digging them out of the box that they are all in

Answer 5

piecing it out shouldnt matter

a'a' aaa or a'a' aab or a'a' abb or a'a' bbb or etc ....

ive got 80 in my possession, i know that 6 are bad and 4 are bad ...

10 out of 80 are bad

whats my error if any?

Answer 6

i see it, i think

70 are good, not 79

Answer 7

10(10*9*(70)*(69)*(68))/(80*79*78*77*76)

.1024

Answer 8

http://www.actuary.com/actuarial-discussion-forum/archive/index.php/t-10406.html?s=7e3604b14480daf9a5f0ca9eb82ad8a4

Answer 9

its a good thing i determined C then lol

Answer 10

:D

Answer 11

320am, brain ws collapsing :) 70 and down are good, not 79

Answer 12

ive got one that is a joint probability function, gonna ask that one next ....and no cheating from nonOS sites lol

Answer 13

http://www.wolframalpha.com/input/?i=%285+choose+2%29+%283%2F8*0.2+%2B+5%2F8*.08%29%5E2+*+%281-%283%2F8*0.2+%2B+5%2F8*.08%29%29%5E3

Answer 14

@amistre64 im getting 0.105
i could be wrong though..

Answer 15

I agree with amistre's work
You can first calculate the probability of
P ( d d d ' d ' d ' ) where d = defective , d ' = non defective, with nonreplacement, and then multiply by 5 choose 2 because
P ( d d d ' d ' d ' ) = P( d' d d d' d' ) = ... you can verify this by making a tree.

Now P(d d d ' d' d' ) = ( 10 * 9 * 70 * 69 * 68)/ ( 80*79*78 *77*76)
, then multiply that by 5 choose 2. and there is confirmation in that link. http://www.actuary.com/actuarial-discussion-forum/archive/index.php/t-10406.html?s=7e3604b14480daf9a5f0ca9eb82ad8a4

I would like to see rational's solution salvaged, that approach could work with some tweaking

Answer 16

that link opens a blank page for me :/

Answer 17

(10/80) * (10/80) * (70/80) * (70/80) * (70/80) * 5 choose 2 = .1046

Answer 18

@rational your solution works "with replacement".