A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is

Options

A. `{:(q,s,s,r):}`
B. `{:(r,s,q,p):}`
C. `{:(q,s,r,p):}`
D. `{:(q,p,r,q):}`

Answer

D

Explanation

Let `E_(i)` denote the event that the bag contains I black and `(12-i)` white balls `(i=0,1,2,…,)` and A denote the event that the four balls drawn are black. Then
`P(E_(i))1/13(i=0,1,2...,12)`
`P((A)/(E_(i)))=(""^(i)C_(4))/(""^(12)C_(4))"for"ige4`
a. `P(A)underset(i=0)overset(12)sumP(E_(i))P((A)/(E_(i)))`
`=1/13xx(1)/(""^(12)C_(4))[""^(4)C_(4)+""^(5)C_(4)+...+""^(12)C_(4)]`
`=(""^(13)C_(5))/(13xx""^(12)C_(4))=1/5`
b. Clearly, `P((A)/(E_(10)))=(""^(10)C_(4))/(""^(12)C_(4))=14/33`
c. By Bayer's theorem,
`P((E_(10))/(A))=(P(E_(10))((A)/(E_(10))))/(P(A))`
`(1/13xx14/33)/(1/5)=70/429`
d. Let B denote the probability of drawing 2 white and 2 black balls. Then
`P((B)/(E_(i)))=0if i=0,1or11,12`
`P((B)/(E_(i)))=(""^(i)C_(2)xx""^(12-i)C_(2))/(""^(12)C_(4))"for"i =2,3,...,10`
`P(B)=underset(i=0)overset(12)sumP(E_(i))P((B)/(E_(i))`
`=1/13xx(1)/(""^(12)C_(4))[2{""^(2)C_(2)xx""^(3)C_(2)+...+""^(10)C_(2)xx""^(2)C_(2)]`
`=1/13xx(1)/(""^(12)C_(4))[2{""^(2)C_(2)""^(10)C_(2)+""^(3)C_(2)+""^(9)C_(2)+...+""^(5)C_(2)xx""^(7)C_(2)}+""^(6)C_(2)xx""^(6)C_(2)]`
`=1/13xx1/495(1287)=1/5`