If the pdf of a curve X is
`f(x)={{:(x/8,0ltxlt4),(0,"elsewhere"):}`
Then `P(Xlt1) and P(Xge2)` are

Options

A. `1/(16),3/4`
B. `1/4,3/8`
C. `5/8,7/(16)`
D. None of these

Answer

A

Explanation

`P(Xlt1)=P(0ltXlt1)=overset1underset0intf(x)dx=overset1underset0int(x/8)dx`
`=1/8[x^2/2]_0^1=1/8[1/2-0]=1/(16)`
`P(Xge2)=P(2leXlt4)=overset4underset2intf(x)dx`
`=overset4underset2int(x/8)dx=1/8[x^2/2]_2^4`
`=1/(16)[4^2-2^2]=(12)/(16)=3/4`