Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 1.

A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar.14; May ’13,’06).

Solution:

Question 2.

The probability that a person chosen at random is left handed (in hand writing) is 0.1. What is the probability that in a group of ten people there is one, who is left handed? (TS Mar.’16; AP Mar. ’17 ’15)

Solution:

Here n = 10

p = 0.1

q = 1 – p = 1 – 0.1 = 0.9

P(X = 1) = ^{10}C_{1} (0.1)^{1} (0.9)^{10 – 1}

= 10 × 0.1 × (0.9)

= 1 × (0.9)

= (0.9)

Question 3.

A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar.14; May ’13, ’06)

Solution:

Given P(X = 1) = P(X = 2)

Question 4.

is the probability distribution of a random variable X. Find the value of K and the variance of X. ( March 2006) (TS Mar. 17)

Solution:

Sum of the probabilities = 1

0.1 + k + 0.2 + 2k + 0.3 + k = 1

4k + 0.6 = 1

4k = 1 – 0.6 = 0.4

k = \(\frac{0.4}{4}\) = 0.1

Mean = (-2) (0.1) + (-1) k + 0(0.2) + 1(2k) + 2(0.3) + 3k

= -0.2 – k + 0 + 2k + 0.6 + 3k

= 4k + 0.4

=4(0.1) + 0.4

= 0.4 + 0.4

= 0.8

μ = 0.8

∴ Variance = 4(0.1) + 1(k) + 0(0.2) + 1(2k) + 4(0.3) + 9k – μ^{2}

= 0.4 + k + 0 + 2k + 4(0.3) + 9k – μ^{2}

= 12k + 0.4 + 1.2 – (0.8)^{2}

= 12(0.1) + 1.6 – 0.64

= 1.2 + 1.6 – 0.64 .

∴ σ^{2} = 2.8 – 0.64 = 2.16

Question 5.

A random variable X has the following probability distribution. (TS & AP Mar. ‘16)

Find

i) k

ii) the mean and

iii) P(0 < X < 5).

Solution:

Sum of the probabilities = 1

⇒ 0 + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7k^{2} + k = 1

⇒ 10k^{2} + 9k = 1

⇒ 10k^{2} + 9k – 1 = 0

⇒ 10k(k + 1) – 1(k + 1) = 0

⇒ (10k – 1) (k + 1) = 0

iii) P(0 < x < 5)

P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= k + 2k + 2k + 3k

= 8k

= 8\(\frac{1}{10}\)

= \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 6.

The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c^{3}, P(X = 1) = 4c – 10c^{2}, P(X = 2) = 5c – 1

i) Find the value of c

ii) P(X < 1),P(1 < X ≤ 2) and P (0 < X ≤ 3) (AP & TS Mar. 15, 13, ‘11, 07, 05; May ‘11’)

Solution:

P(X = 0) + P(X = 1) + P(X = 2) = 1

3c^{3} + 4c – 10c^{2} + 5c – 1 = 1

3c^{3} – 10c^{2} + 9c – 2 = 0

c = 1 satisfy this equation

c = 1 ⇒ P(X = 0) = 3 which is not possible

Dividing with c – 1, we get

3c^{2} – 7c + 2 = 0

(c – 2) (3c – 1) = 0

ii) P(1 < X ≤ 2) = P(X = 2) = 5c – 1

= \(\frac{5}{3}\) – 1 = \(\frac{2}{3}\)

iii)P(0 < X ≤ 3) = P(X = 1) + P(X = 2)

= 4c – 10c^{2} + 5c – 1

= 9c – 10c^{2} – 1

= 9.\(\frac{1}{3}\) – 10.\(\frac{1}{9}\) – 1

= 3 – \(\frac{10}{9}\) – 1 = 2 – \(\frac{10}{9}\) = \(\frac{8}{9}\)

Question 7.

One in 9 ships is likely to be wrecked, when they are set on sail, when 6 ships are on sail, find the probability for

i) Atleast one will arrive safely

ii) Exactly, 3 will arrive safely. (Mar. 2008)

Solution:

p = probability of ship to be wrecked = \(\frac{1}{9}\)

Question 8.

If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X ≤ 4). (May ’06)

Solution:

Mean = np = 2.4 …… (1)

Variance = npq = 1.44 …… (2)

Dividing (2) by (1),

\(\frac{\mathrm{npq}}{\mathrm{np}}\) = \(\frac{1.44}{2.4}\)

q = 0.6 = \(\frac{3}{5}\)

2

p = 1 – q = 1 – 0.6 = 0.4 = \(\frac{2}{5}\)

Substituting in (1)

Question 9.

The probability distribution of a random variable X is given below. (AP Mar. ‘17’) (Mar. ‘14; May ‘13)

Find the value of K, and the mean and variance of X.

Solution:

Question 10.

The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X ≥ 1)

(AP Mar. ‘16 TS Mar. 17 ‘15, ’08)

Solution:

Given distribution ¡s Binomial distribution with mean = np = 4

variance = npq = 3

∴ \(\frac{n p q}{n p}\) = \(\frac{3}{4}\)

⇒ q = \(\frac{3}{4}\)

so that p = 1 – q

= 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)

∴ np = 4

n\(\frac{1}{4}\) = 4

⇒ n = 16

Question 11.

A cubical die is thrown. Find the mean and variance of X, giving the number on the face that shows up.

Solution:

Let S be the sample space and X be the random variable associated with S, where P(X) is given by the following table

Question 12.

The probability distribution of a random variable X is given below.

Find the value of k, and the mean and variance of X

Solution:

Question 13.

If x is a random variable with probability distribution. P(X = k) = \(\frac{(k+1) c}{2^{k}}\), k = 0, 1, 2 then find c.

Solution:

Question 14.

Let X be a random variable such that P(X = -2) =P(X = -1) = P(X = 2) = P(X = 1) = \(\frac{1}{6}\) and P(X = 0) = \(\frac{1}{3}\). Find the mean and variance of X.

Solution:

Question 15.

Two dice are rolled at random. Find the probability distribution of the sum of the numbers on them. Find the mean of the random variable.

Solution:

When two dice are rolled, the sample space

S contains 6 × 6 = 36 sample points.

S = {(1, 1), (1, 2) (1, 6), (2, 1), (2, 2) (6, 6)}

Let X denote the sum of the numbers on the tw0 dice.

Then the range of X = {2, 3, 4, ……… 12}

The Prob. distribution of X is given by the following table.

Question 16.

8 coins are tossed simultaneously. Find the probability of getting atleast 6 heads.

Solution:

p = Probability of getting head = \(\frac{1}{2}\)

q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) ; n = 8

P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)

Question 17.

The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X ≥ 1)

(A.P. Mar. ’16, T.S. Mar. ’15, ’08)

Solution:

Given distribution is Binomial distribution with mean = np = 4

variance = npq = 3

∴ \(\frac{n p q}{n p}\) = \(\frac{3}{4}\)

⇒ q = \(\frac{3}{4}\)

so that p = 1 – q

= 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)

∴ np = 4

n\(\frac{1}{4}\) = 4

⇒ n = 16

Question 18.

The probability that a person chosen at random is left handed (in hand writing) is 0.1. What is the probability that in a group of ten people there is one, who is left handed ? (Mar. 16, AP. Mar. ’15)

Solution:

Here n = 10

p = 0.1

q = 1 – p = 1 – 0.1 = 0.9

P(X = 1) = ^{10}C_{1} (0.1)^{1} (0.9)^{10 – 1}

= 10 × 0.1 × (0.9)

= 1 × (0.9)

= (0.9)

Question 19.

In a bõok of 450 pages, there are 400 typographical errors. Assumiñg that the number of errörs per page follow the

poisson law, find the probability that a random sample of 5 pages will contain no typographical error.

Solution:

The average number of errors per page in the book is

The required probability that a random sample of 5 pages will contain no error is

[P(X = 0)]^{5} = \(\left(e^{-8 / 9}\right)^{5}\)

Question 20.

Deficiency of red cells in the blood cells is determined by examining a specimen of blood under a microscope. Suppose a small fixed volume contains on an average 20 red cells for normal persons. Using the poisson distribution, find the probability that a specimen of blood taken from a normal person will contain less than 15 red cells.

Solution:

Question 21.

A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar. ‘14; May ‘06, ‘13)

Solution:

Given P(X = 1) = P(X = 2)