Home
Class 12
MATHS
The p.d.f. of a continuous r.v. X is f(x...

The p.d.f. of a continuous r.v. X is `f(x)={{:((1)/(10)","-5lexle5),(0", otherwise"):}`, then `P(X lt0)` =

A

`(1)/(2)`

B

`(1)/(10)`

C

`(2)/(5)`

D

`(1)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • PROBABILITY

    MODERN PUBLICATION|Exercise EXERCISE 13 (f) (LATQ I)|10 Videos

  • PROBABILITY

    MODERN PUBLICATION|Exercise EXERCISE 13 (f) (LATQ II)|5 Videos

  • PROBABILITY

    MODERN PUBLICATION|Exercise EXERCISE 13 (e) (LATQ)|14 Videos

  • MATRICES

    MODERN PUBLICATION|Exercise CHAPTER TEST (3)|12 Videos

  • RELATIONS AND FUNCTIONS

    MODERN PUBLICATION|Exercise CHAPTER TEST (1)|12 Videos

Similar Questions

Explore conceptually related problems

The p.d.f. of a continuous r.v. X is f(x)={{:(k(4-x^(2))","-2lexle2),(0", otherwise"):} , then P(Xgt0) =

The p.d.f. of a continuous r.v. X is f(x)={{:((x^(2))/(3)","-1ltxlt2),(0", otherwise"):} , then P(X lt 1) =

The p.d.f. of a continuous r.v. X is f(x)={{:(k(4-x^(2))","-2lexle2),(0", otherwise"):} , then k =

The p.d.f. of a continuous r.v. X is f(x)={{:((x)/(8)","0ltxlt4),(0", otherwise"):} , then F (5) =

The p.d.f. of a continuous r.v. X is f(x)={{:((x)/(8)", "0ltxlt4),(0", otherwise"):} , then F (x) =

The p.d.f. of continuous r.v. X is f(x)={{:((x)/(8)","0ltxlt4),(0", otherwise"):} , then P(Xle2) =

The p.d.f. of continuous r.v. X is f(x)={{:((x)/(8)","0ltxlt4),(0", otherwise"):} , then P(Xgt3) =

The p.d.f. of a continuous r.v. X is f(x)={{:(k(4-x^(2))","-2lexle2),(0", otherwise"):} , then P(Xlt-0.5" or "Xgt0.5) =

The p.d.f. of a continuous r.v. X is f(x)={{:((x)/(8)","0ltxlt4),(0", otherwise"):} , then F (0.5) =

The p.d.f. of a continuous r.v. X is f(x)={{:((x^(2))/(3)","-1lt x lt 2),(0", otherwise"):} , then P(X gt0) =