Question Bank

Unit-2

Question-1: If a continuous random variable X has the following probability density function:

Then find-

1. P[0.2 < X < 0.5]

Solution:

Here f(x) is the probability density function, then-

Question-2: The pdf of a continuous random variable is given as below-

Then find the c.d.f. Of X.

Solution:

The c.d.f. Of X will be given by-

Hence the c.d.f will be-

Question-3: :Following probability distribution

x	0	1	2	3	4	5	6	7
P(x)	0

Find: (i) k   (ii)

(i)                Distribution function

(ii)              If   find minimum value of C

(iii)           Find

Solution:

If P(x) is p.m.f –

(i)

X	0	1	2	3	4	5	6	7
P(x)	0

(ii)

(iii)

(iv)

(v)

Question-4: Let be two independent N(0,1) random variables. Define

Where is a real number in (-1,1)

a. Show that X and Y are bivariate normal.

b. Find the joint PDF of X and Y.

c. Find (X,Y)

Sol. First, note that since are normal and independent , they are jointly normal with the joint PDF

We need to show aX+bY is normal for all we have

and thus it is normal.

We can use the methods of transformations (Theorem 5.1) to find the joint PDF of X and Y . The inverse transformation is given by

Thus we conclude that,

To find , first mode

Therefore,

Question-5: Let X and Y be jointly normal random variables with parameters Find the conditional distribution of Y given X =x.

Solution:

One way to solve this problem is by using the joint PDF formula since X N we can use

Thus given X=x, we have

And,

Since are independent, knowing does not provide any information on . We have shown that given X=x,Y is a linear function of , thus it is normal. Ln particular

We conclude that given X=x,Y is normally distributed with mean and variance
 

Question-6: Let X and Y be jointly normal random variables with parameters
 

Find

Find

Find

Solution:

a. Since X and Y are jointly normal, the random variable V-2X+Y is normal. We have

Thus, , therefore,

Note that

Using Properties, we conclude that given X=2,Y is normally distributed with

Thus,

Question-7: A die is tossed thrice. A success is getting 1 or 6 on a toss. Find the mean and variance of the number of successes.

Solution:

Probability of mean success =

Probability of failures=

Probability of no success =Probability of all 3 failures =

Probability of one success and 2 failures

Probability of two success and one failure

Probability of three successes

Now,

0            1            2          3

=8/27       4/9        2/9       1/27

Mean,

Also,
 

Variance,
 

Question-8: The probability density function of a variate X is

X          0           1           2             3              4             5             6

P(X)     k          3k         5k           7k          9k             11k        13k

Solution:

Thus minimum value of k

Question-9: A random variable X has the following probability function

x:               0       1        2           3        4       5      6        7

p(x)          0        k       2k         2k      3k    

 Solution:

 (i) If X is a random variable then

i.e.
 

(ii)
 

(iii)
 

Question-10: A bag contains 12 pens of which 4 are defective. Three pens are picked at random from the bag one after the other.

Then find the probability that all three are non-defective.

Solution:

Here the probability of the first which will be non-defective = 8/12

By the multiplication theorem of probability,

If we draw pens one after the other then the required probability will be-

Question-11: The probability of A hits the target is 1 / 4 and the probability that B hits the target is 2/ 5. If both shoot the target then find the probability that at least one of them hits the target.

Solution:

Here it is given that-

Now we have to find-
 

Both two events are independent. So that-

Question-12: An urn I contains 3 white and 4 red balls and an urn II contains 5 white and6 red balls. One ball is drawn at random from one of the urns and is found to be white.

Find the probability that it was drawn from urn I.

Solution:

Let

Now we have to find-

So that by using Bayes theorem, we get-

Here,

Two urns are equally likely to select,

So that-