Probability Density Function

Question

Comments

@Habibkhan please help

Answer 1

i think answer should be A.though i am not geting the xact answer.

P(X>b|X<b/2) = P(X>b and X< b/2)/ P(X<b/2)

now,it is given that -1 < b < 0

so, -1/2  < b/2<  0

hence P(X< b/2) = integral [3x²]^b/2 _-1/2 = [x³] ^b/2_-1/2 = b³ /8 + 1/8

and P(X > b and X< b/2 ) = [x³] ^b/2_b =-7b³/8

hence,answer is -7b³/b³ + 1

so might be i am wrong somewhere and answer is A.

please confirm

Comments

I am getting answer D.

 

P(X<b/2) = integrate from -1(lower bound)  to b/2(upper bound)

I have taken -1 as lower bound because we are considering values of X which has least value of -1.

 

P(X>b and X<b/2) = integrate from b(lower bound)  to b/2(upper bound)

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@sushant,for b ,it is -1,but for b/2,we have to divide by 2 on both sides..otherwise how will the equality remain..??

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@Akriti. Thats what I am saying. Even though they have X<b/2, they havent said X>b/2.

We are integrating  X and hence, when X<b/2, X can even take value -1.