probability(2) 20點

Denote
A be the event Bin 1 was chosen,
B be the event Bin 2 was chosen, and
C be the event 2 of the 3 parts obtained are defective.
P(A and C)
= 1/2 * (5C2 * 15C1) / 20C3
= 1/2 * 150 / 1140
= 0.06579
P(C)
= P(A and C) + P(B and C)
= 1/2 * (5C2 * 15C1) / 20C3 + 1/2 * (4C2 * 11C1) / 15C3
= 1/2 * 150 / 1140 + 1/2 * 66 / 455
= 0.1383
P(A | C) = P(A and C) / P(C) = 0.4756 (4 d.p.)

P(2 of 3 parts obtained are defective in bin 1)
= 3C2 x 5/20 x 4/19 x 15/18
= 3 x 1/4 x 4/19 x 5/9
= 5/57

P(2 of 3 parts obtained are defective in bin 2)
= 3C2 x 4/15 x 3/14 x 11/13
= 3 x 2/5 x 1/7 x 11/13
= 66/455

Then, P(bin 1 is chosen | 2 of the 3 parts obtained are defective)
= P(bin 1 is chosen and 2 of the 3 parts obtained are defective) / P(2 of the 3 parts obtained are defective)
= (5/57) / (5/57 + 66/455)
= 0.3768

There are 2 parts which are defective in the sample. There is only 1 part which is not defective.
There are 5 parts in Bin 1 which are defective but only 2 parts are selected from Bin 1 . There are 15 parts in Bin 1 which are not defective. If Bin 1 is chosen, there are 5C2*15C1=150 combinations.
There are 4 parts in Bin 2 which are defective but only 2 parts are selected from Bin 2 . There are 11 parts in Bin 2 which are not defective. If Bin 2 is chosen, there are 4C2*11C1=66 combinations.
Probability that Bin 1 was chosen
=150/(150+66)
=0.6944444...
=69.44%(corr. to 2 decimal place)

As it is possible to have 2 of the 3 parts abtained are defective for both 2 bins,
The probability that Bin 1 was chosen = 1/2