Joint Probability Density Function MCQ Quiz - Objective Question with Answer for Joint Probability Density Function - Download Free PDF

Last updated on Apr 14, 2025

Latest Joint Probability Density Function MCQ Objective Questions

Joint Probability Density Function Question 1:

Two random variables X and Y are distributed according to

fX,Y(x,y)={(x+y)0x10y10,otherwise.

The probability (X+Y1) is ________

Answer (Detailed Solution Below) 0.33

Joint Probability Density Function Question 1 Detailed Solution

Concept: fxy(x,y)={(x+y),0x1,0y10,otherwiseis given as joint pdf.  The probability of the required region can be found by integrating the joint pdf over the given region. 

Application: We have, fxy(x,y)={(x+y),0x1,0y10,otherwise

Now, region x+y1 implies

y1x

plotting the region in xy plane we have

Gate EC 2016 paper 2 Images-Q28

The rectangle bounded by 0x1 and 0y1 defined the region for which random variables x and y are defined. The shaded region shows the region x+y1.

Now,

P(X+Y1)=y=01x=01y(x+y)dxdy=y=01((1y)22+y(1y))dy=12y=01[(1y)[1y+2y]]dy=12y=01(1y2)dy=12[yy33]01=12[113]P(X+Y1)=13

Joint Probability Density Function Question 2:

The joint pdf of a bivariate random variable (X,Y) is given by,

fXY(x,y)={kxy0<x<1,0<y<10otherwise

where K is a constant. The value of K is _______.

Answer (Detailed Solution Below) 4

Joint Probability Density Function Question 2 Detailed Solution

The range space RXY is shown below.

Gate EC Communication part test Images Q3

Since, fXY(x,y) is a pdf

0101fXY(x,y)dxdy=1

K0101xydxdy=1

K01y(x22|01)dy=1

K01y2dy=1

K/4=1

K=4

Top Joint Probability Density Function MCQ Objective Questions

Two random variables X and Y are distributed according to

fX,Y(x,y)={(x+y)0x10y10,otherwise.

The probability (X+Y1) is ________

Answer (Detailed Solution Below) 0.33

Joint Probability Density Function Question 3 Detailed Solution

Download Solution PDF

Concept: fxy(x,y)={(x+y),0x1,0y10,otherwiseis given as joint pdf.  The probability of the required region can be found by integrating the joint pdf over the given region. 

Application: We have, fxy(x,y)={(x+y),0x1,0y10,otherwise

Now, region x+y1 implies

y1x

plotting the region in xy plane we have

Gate EC 2016 paper 2 Images-Q28

The rectangle bounded by 0x1 and 0y1 defined the region for which random variables x and y are defined. The shaded region shows the region x+y1.

Now,

P(X+Y1)=y=01x=01y(x+y)dxdy=y=01((1y)22+y(1y))dy=12y=01[(1y)[1y+2y]]dy=12y=01(1y2)dy=12[yy33]01=12[113]P(X+Y1)=13

Joint Probability Density Function Question 4:

Two random variables X and Y are distributed according to

fX,Y(x,y)={(x+y)0x10y10,otherwise.

The probability (X+Y1) is ________

Answer (Detailed Solution Below) 0.33

Joint Probability Density Function Question 4 Detailed Solution

Concept: fxy(x,y)={(x+y),0x1,0y10,otherwiseis given as joint pdf.  The probability of the required region can be found by integrating the joint pdf over the given region. 

Application: We have, fxy(x,y)={(x+y),0x1,0y10,otherwise

Now, region x+y1 implies

y1x

plotting the region in xy plane we have

Gate EC 2016 paper 2 Images-Q28

The rectangle bounded by 0x1 and 0y1 defined the region for which random variables x and y are defined. The shaded region shows the region x+y1.

Now,

P(X+Y1)=y=01x=01y(x+y)dxdy=y=01((1y)22+y(1y))dy=12y=01[(1y)[1y+2y]]dy=12y=01(1y2)dy=12[yy33]01=12[113]P(X+Y1)=13

Joint Probability Density Function Question 5:

The joint pdf of a bivariate random variable (X,Y) is given by,

fXY(x,y)={kxy0<x<1,0<y<10otherwise

where K is a constant. The value of K is _______.

Answer (Detailed Solution Below) 4

Joint Probability Density Function Question 5 Detailed Solution

The range space RXY is shown below.

Gate EC Communication part test Images Q3

Since, fXY(x,y) is a pdf

0101fXY(x,y)dxdy=1

K0101xydxdy=1

K01y(x22|01)dy=1

K01y2dy=1

K/4=1

K=4

Joint Probability Density Function Question 6:

Let X and Y be two Gaussian random variables with mean μx = 6, μy = -2 and variable σx = σy = 4 respectively. The respected value of XY is E[XY] = -22. If a random variable is defined as Z = X + 3Y, what will be the probability density function of Z?

  1. 1200πeZ2/200
  2. 1200πeZ2/200
  3. 12πeZ2/2
  4. 12πeZ2/2

Answer (Detailed Solution Below)

Option 1 : 1200πeZ2/200

Joint Probability Density Function Question 6 Detailed Solution

X̅ = μx = 6, Y̅ = μy = -2

The variance of the random variables is given as σx = σy = 4

The expected value is 22, i.e.

E[XY]=XY=22

Random variable is defined as Z = X + 3Y

Since X and Y are Gaussian random variables and Z is a linear combination of the two, Z will be also a Gaussian random variable.

The mean of random variable Z is obtained as:

Z¯=X+3Y=X¯+3Y¯

= 6 + 3 × (-2)

= 0

Again, the variance of Z is obtained as:

σZ2=Z¯2Z¯2=Z¯2

=(X+3Y)2=X2+9Y2+6XY

=X¯2+9Y¯2+6XY

=(σX2+X¯2)+9(σY2+Y¯2)+6XY

= (16 + 36) + 9(16 + 4) + 6 (-22) = 100

Thus, the PDF of Z is given by:

fZ(Z)=12πσZ2e(ZmZ)22σZ2

=1200πe(Z0)22×100

fz(z)=1200πeZ2e200

Joint Probability Density Function Question 7:

Let x, y, z be independent and identical random variable uniformly distributed on [0, 1] probability P[x + y ≤ z] is _______

Answer (Detailed Solution Below) 0.15 - 0.17

Joint Probability Density Function Question 7 Detailed Solution

P[x + y ≤ z]

⇒ P[x + y – z ≤ 0]

Let m = x + y – z

fm(x) = fx(x) * fy(x) * fz(-x)

Communication system file 1 images Q30

Communication system file 1 images Q30a

Communication system file 1 images Q30b

fm(x)=[r(x)2r(x1)+r(x2)][u(x+1)u(x)]

Using differentiation property

fm(x)=[4(x)24(x1)+4(x2)][u(x+1)u(x)]

=r(x+1)r(x)2r(x)+2r(x1)+r(x1)r(x2)

fm(x)=r(x+1)3r(x)+3r(x1)r(x2)

Integrating

fm(x)=(x+1)22u(x+1)32x2u(x)+32(x1)2u(x1)+12(x2)2u(x2)

P(M0)=10fm(x)dx

=10(x+1)22dx=[12(x+1)33]10=16

= 0.167

Joint Probability Density Function Question 8:

A joint probability density function of the random variable x, y, z is

F(x, y, z) = 8 xyz , 0 < x, y, z < 1 then find P(x < y < z) (Upto 2 decimal places)

Answer (Detailed Solution Below) 0.16 - 0.18

Joint Probability Density Function Question 8 Detailed Solution

Since each random variable has some type of distribution each of them has an equal probability of being the highest,

 Either x can be highest and y and z are small or 

 y can be highest and x and y are smaller 

 z can be largest and x and y are smaller.

So all three have an equal probability of being largest.

Hence out of all,  z is largest is 1/3 

if z is the largest other 2 cases are

x > y 

y >x 

Both have equal probability

Hence P(x < y < z) = 12(13) = 0.166

Joint Probability Density Function Question 9:

The joint pdf of X and Y is given by

fXY(x,y)=kce(ax+by)u(x)u(y) where a,b and c are constants. What is the value of k?

u(x) and u(y) are unit step functions

  1. (a+b)c
  2. abc
  3. (a+b)/c
  4. ab/c

Answer (Detailed Solution Below)

Option 2 : abc

Joint Probability Density Function Question 9 Detailed Solution

Since fXY(x,y) is joint pdf

fXY(p,q)dpdq=1kce(ap+bq)dpdq=kc0eapdp0ebqdqfXY(p,q)dpdq=kc1a1b=1k=abc

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