Important 2 marks Questions with Answers

Important '2 Marks' Questions and Answers

Type 1. Probability - Axioms of probability - Conditional probability Baye's theorem - Discrete and continuous random variable - Moments - m.g.f

1. Define random variable. [A.U CBT N/D 2011]

A random variable is a function that assigns a real number to each outcome in the sample space for random experiment.

2. Define discrete random variable with an example. [A.U CBT A/M 2011] [A.U A/M 2019 (R13) (RP)]

A random variable whose set of possible values is either finite or countably infinite is called discrete.

Example : number of transmitted bits received in error.

3. Define continuous random variable with an example. [A.U A/M 2019 (R13) (RP)]

A random variable X is said to be continuous if it takes all possible values between certain limits say from real number 'a' to real number 'b'.

Example: length, pressure, temperature, time, voltage, weight.

4. Define : Probability mass function :

For a discrete random variable X with possible values x1, x2, ..., xn a probability mass function is a function such that

5. Define cumulative distribution function of a discrete random variable X.

Solution :

The cumulative distribution function of a discrete random variable X, denoted as F (x), is

For a discrete random variable X, F(x) satisfies the following properties.

i. F(x) = P(X = x) = 

ii. 0 = F(x) = 1

iii. If x = y, then F(x) = F(y).

6. Define mean (or) expected value of a discrete random variable.

Solution:

The mean or expected value of the discrete random variable X, denoted as µ or E(X) is

The variance of X, denoted as σ² or V(X) is

7. Probability density function f(x) can be used to describe the probability of a ........... random variable X.

Solution: Continuous.

8. Define probability density function. [p.d.f]

Solution: 

9. Define cumulative distribution function of a continuous random variable X.

Solution: 

10. Define mean or expected value of a continuous random variable X.

Solution : 

11. Test whether f(x) =  can be the probability density function of a continuous random variable. [A.U N/D 2014] [A.U A/M 2015 R13]

Solution

Given: f(x) = 

To prove: 

12. Define expectation of a random variable.

Solution :

Consider a random variable X with p.m.f. p (x) (or) p.d.f. f(x). If g(X) is function of X, then g(X) is a random variable. If E [g (x)] exists then

13. Define n^th moment about origin for a random variable.

Solution :

The nth moment about origin of a random variable X is defined as the expected value of the nth power of X.

For discrete random variable X,

For continuous random variable X

14. Define moment generating function.

Solution :

Def : Moment generating function of a random variable X about the origin is defined as

where t being a real parameter assuming that the integration or summation is absolutely convergent for some positive number h such that |t| < h.

15. If the p.d.f of a random variable X is f(x) = x/2 in 0 ≤ x ≤ 2, find P (X > 1.5 / X > 1). [A.U A/M 2010]

Solution :

Given : f(x) = x/2, 0 ≤ x ≤ 2

16. The c.d.f of a continuous random variable is given by 

Find the p.d.f and mean of X. [A.U A/M 2011, N/D 2011]

Solution:

17. Find C, if P [X = n] = C (2/3)ⁿ ; n = 1, 2, ......... [[A.U M/J 2012]

Solution:

We know that, P[X = n] = 1 => 

18. An experiment succeeds twice as often as it fails. Find the chance that in the next 4 trials, there shall be atleast one success. [A.U N/D 2012]

Solution :

p = 2/3, q = 1/3, n = 4

19. A random variable X has cdf  Find the p.d.f of X and the expected value of X. [AU M/J 2013, M/J 2014]

Solution:

Given:

 

20. Suppose that, on an average, in every three pages of a book there is one typograhpical error. If the number of typographical errors on a single page of the book is a Poisson random variable. What is the probability of at least one error on a specific page of the book? [A.U A/M 2015 R-8] [A.U A/M 2018 R-8]

Solution :

21. A continuous random variable X has a p.d.f given by Find P(X > 1) [A.U N/D 2015 R-8, R-13]

Solution :

Given: 

22. Show that the function f(x) =  is a probability density function of a random variable X. [A.U A/M 2015 (RP) R-13]

Solution:

Given: f(x) = 

23. If f (x) =  is the p.d.f of a random variable X, then the value of K. [A.U M/J 2016 R-13 RP]

Solution:

We know that, 

24. Find the expected value of the discrete random variable X with the probability mass function p (x) =  [A.U N/D 2016 R-13 R8]

Solution :

25. A random variable X is known to have a distribution function , where b > 0 is a constant. Determine its density function. [A.U_N/D 2016 R13 RP]

Solution :

26. Find the value of k for a continuous random variable x whose density function is given by f(x) = k x² e^-x, x ≥ 0 [A.U N/D 2018 R-17] [PS]

Solution :

27. Let X be the random variable which denotes the number of heads in three tosses of a fair coin. Determine the probability mass function of X. [A.U N/D 2015 R-8] [A.U A/M 2019 R13 (PQT)]

Solution :

28. Let A and B be two events such that P(A) = 0.5, P(B) = 0.3 and P (A∩B) = 0.15 compute P (B/A) and  [A.U N/D 2019 (R17) (RP)]

Solution :

Given: P(A) = 0.5, P(B) = 0.3, P(A∩B) = 0.15

29. Let A and B be two events such that P(A) = 1/3, P(B) = 3/4 and P (A∩B) = 1/4. Compute P (A/B) and  [A.U A/M 2019 (R17) (PQT)]

Solution :

30. A bag contains 8 white and 4 black balls. If 5 balls are drawn at random, what is the probability that 3 are white and 2 are black? [A.U N/D 2019 (R17)-PQT]

Solution :

Total number of balls = 8 + 4 = 12

S = {5 balls are taken out of 12}

The number of favourable out comes

(i.e., 3 are white and 2 are black)

31. Let M_X(t) = 1/1-t =, |t| < 1, the moment generating function of R.V.X. FInd E [X] and E [X²] [A.U N/D 2019 (R17) - PQT]

Solution :

32. A random variable X has probability mass function P(X = x) = x/10, x = 1, 2, 3, 4. Find the cumulative distribution function, F(x) of X. [A.U. A/M 2019 (R17) PQT]

Solution :

33. The R.V.X has p.m.f P (X = x) = Obtain: (i) The value of 'c', (ii) P (X ≥ 2) [A.U N/D 2019 (R17) R.P]

Solution :

34. Show that for any events A and B in S, P(B) = P(B/A) P (A) +  [A.U A/M 2019 (R17) R.P]

Solution :

35. The moment generating function of a random variable X is given  What is P [X = 0]? [A.U A/M 2019 (R8) RP]

Solution :

36. The probability density function of the random variable X is given by 

Find the value of k. [A.U A/M 2019 (R17) PS]

Solution :

37. If A and B are mutually exclusive events P(A) = 0.29 and P(B) = 0.43 then find P(Ā) and P (AUB) [A.U N/D 2019 (R17) PS]

Solution :

Given: 

Type 2. B.D, P.D, G.D, U.D, E.D, N.D

B.D. 1. Define Binomial Distribution.

A random variable X is said to follow Binomial distribution if it assumes only non-negative values and its probability mass function is given by

B.D. 2. Define Binomial frequency distribution.

Let us suppose that n trials constitute an experiment. Then if this experiment is repeated N times, the frequency function of the binomial distribution is given by,

The expected frequencies of 0, 1, 2, n successes are given by he successive terms of N (q + p)ⁿ.

B.D. 3. For a Binomial distribution mean is 6 and S.D. is v2. Find the first two terms of the distribution. [A.U. A/M 2004, M/J 2014]

Solution:

For a Binomial distribution, mean = np = 6 (given) and variance = npq

The probability mass function of a Binomial distribution is given by

Hence, the first two terms are given by,

B.D. 4. The mean and variance of binomial distribution are 5 and 4. Determine the distribution. [A.U A/M 2015 R-13]

Solution: 

Binomial distribution : 

Given (mean) np = 5, (variance) npq = 4

B.D.5. Let X be a random variable with moment generating function M_x (t) = 

then find its mean and variance. [A.U M/J 2016 R13 RP]

Solution :

Given 

This is of the form , which is the m.g.f of binomial distribution.

B.D. 6. If the probability of success is 1/100, how many trials are necessary in order that the probability of atleast one success is greater than 1/2? [A.U N/D 2016, R-13 RP]

Solution:

=> 68.4 < n

=> n = 69

B.D.7. An experiment succeeds twice as often as it fails. Find the chance that in the next 4 trails, there shall be atleast one success. [A.U A/M 2019 (R8) R.P]

Solution :

Let X -> Number of successes

Since we are talking about success and failure.

It is a Bernoulli trial.

So, X has a binomial distribution.

n = number of trials = 4

p = Probability of success

q = Probability of failure = 1 - p

Given: Experiment succeeds twice as often as it fails.

Probability of atleast one success = P[X ≥ 1] = 1 - P[X < 1] = 1 - P[X = 0]

B.D.8. For a binomial distribution mean is 2 and variance is 4/3, find the first term of the distribution. [A.U A/M 2019 (R17) PS]

Solution :

 

P.D. 1. Define Poisson Poisson distribution and state any two instances where Poisson distribution may be successfully employed. [A.U N/D 2006]

Solution:

Poisson Distribution: A random variable X is said to follow Poisson distribution if it assumes only non-negative values and its probability mass function is given by

where, λ is known as the parameter of Poisson distribution.

State any two instances where Poisson distribution may be successfully employed.

i. Number of printing mistakes at each page of the book.

ii. Number of suicides reported in a particular day.

iii. Number of deaths due to a rare disease.

iv. Number of defective items produced in the factory.

P.D.2. Define Poisson frequency distribution.

Solution :

P.D.3. What are the limitations of Poisson distribution? [A.U A/M 2015 - R13]

Solution:

(i) n is indefinitely large (i.e.,) n -> ∞

(ii) p is very small s.t. p -> 0

(iii) np = λ (a finite quantity)

=> p = λ /n and q = 1 - p = 1 - λ/n

( G.D.1.) Define Geometric distrbution.

Suppose that independent trials, each having a probability p, 0 < p < 1, of being a success, are performed until a success occurs. If we get X equal the number of trials required, then

G.D.2. State memory less property and which continuous and discrete distributions follow this property. [A.U A/M 2019 (R13) PQT]

Solution :

If X is a random variable (discrete or continuous), then for any two positive integers m and n.

P[X > m+n / X > m] = P[X > n] which is the memoryless property.

Discrete: Geometric distribution.

Continuous: Exponential distribution.

G.D.3. Find the second moment about the origin of the Geometric distribution with parameter p. [A.U A/M 2019 (R17) R.P]

Solution :

U.D. 1. If 'X' is Uniformly distributed in (-π/2, π /2), find the probability distribution function of y = tan x. [A.U. N/D 2003, 2010]

Solution.

Since, 'X' is Uniformly distributed in (-π/2, π/2), its pdf is

f(x) = 1/ π

We first determine the distribution function of Y.

F(y) = P(Y≤ y) = P (tan X ≤ y) = P(X ≤ tan^-1 y)

U.D. 2. If the m.g.f of a Uniform distribution for a random variable X is , find E (X). [A.U A/M 2010]

Solution: 

Given: 

In Uniform distribution: 

We know that, 

From (1) & (2) we get, b = 5, a = 4

U.D. 3. A random variable X is uniformly distributed between 3 and 15. Find the variance of X. [A.U N/D 2015 [R13, R8]

Solution:

Given: a = 3, b = 15

Variance 

E.D. 1. Define exponential distribution.

A continuous random variable X is said to follow Exponential distribution if its probability density function is given by,

E.D. 2. The time (in hours) required to repair a machine is Exponentially distributed with parameter λ = 1/2. What is the probability that a repair takes atleast 10 hours given that its duration exceeds 9 hours ? [A.U. N/D 2004]

Solution:

If X represents the time to repair a machine which is Exponentially distributed, the density function of X is,

P(the repair takes atleast 10 hours given that its duration exceeds P(X  ≥ 10 / X > 9)

= P(X > 1) by the memoryless property.

(N.D. 1.) Find the mean of a Normal distribution. [A.U N/D 2018, R17, PS]

Solution :

Mean = E [X] = µ