A bag contains some white and some black balls, all combinations being equally likely. The total number of balls in the bag is 12. Four balls are drawn at random from the bag at random without replacement.

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`