probability

[Var1995]

do we have to take values for the no. of defective balls to solve this?

[Rhythm]

Answer is (B):

                                  NON DEFECTIVE(N)                         DEFECTIVE(D)                p(D)
URN 1                                 9                                             1                              1/10
URN 2                                 8                                             2                              2/10
URN 3                                 7                                             3                              3/10
URN 4                                 6                                             4                              4/10
URN 5                                 5                                             5                              5/10

Defective balls- 15
Total balls- 50
Probability of defective balls= 15/50 =  3/10

For part B, use Bayes theorem.

[Var1995]

on what basis u have taken these values?

[Ansh malhotra]

You are right it is not written in the question that urn I contains I defective balls

[urvashi]

No bayes theorem will not be correct

[Var1995]

there is printing mistake, it is urn i contains i defective balls....

[Var1995]

bayes' thm will be used as follows:

let the events be:
A= ball is defective
E1= ball is from urn 1
E2= urn 2
E3= urn 3
E4= urn 4
E5= urn 5

p(E2/A)= {p(A/E2) p(E2)}/ {p(A/E1) p(E1)+  p(A/E2) p(E2)+  p(A/E3) p(E3)+ p(A/E4) p(E4)+ p(A/E5) p(E5)}