Differentiate `y = (x − cos^2x)^4`

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`