Students must practice this TS Intermediate Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a)

I.

Question 1.

A p.d.f of a discrete random variable h zero except at the point to x = 0, 1, 2. At these points it has the value P(0) = 3c^{3} P(1)= 4c – 10c^{2}, P(2)= 5c – 1 for same c > 0. Find the value of c.

Solution:

Given P(0) = 3c^{3},

P(1) = 4c – 10c^{2},

P(2) = 5c – 1

we have \(\sum_{i=0,1,2}\) P(i) = 1

⇒ P(0) + P(1) + P(2) = 1

⇒ 3c^{3} + 4c – 10c^{2} + 5c – 1 = 1

⇒ 3c^{3} – 10c^{2} + 9c – 2 = 0

c = 1 satisfy the equation by inspection and for c = 1, we have

P(0) = 3,

P(1) = 4 – 10 = 6,

P(2) = 4

Hence c = 1 does not satisfy 0 ≤ P (E) ≤ 1.

So we try for other solutions by synthetic division method.

∴ 3c^{2} – 7c + 2 = 0

⇒ 3c^{2} – Gc – c . 2 0

⇒ 3c (c – 2) – 1 (c – 2 )= 0

(3c – 1) (c – 2 ) = 0

c – 2 = 0

⇒ c = 2 is not admissible;

∴ c = \(\frac{1}{3}\) suit the solutions P (0), P (1) and P (2)

∴ c = \(\frac{1}{3}\).

Question 2.

Find the constant c, so that F(x)= c\(\frac{2}{3}\), x = 1, 2, 3, …………… is the p.d.f. of a discrete random variable X.

Solution:

Since F (x) is the p.d.f. oF discrete random variable x.

we have \(\sum_{x=1}^{\infty}\) F(x) = 1

Question 3.

is the probability distribution of a random variable X. Find the value of k and the variance of X.

Solution:

Sum of the probability = 1

⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1

⇒ 4k + 0.6 = 1

⇒ 4k = 0.4

⇒ k = 0.1

Mean = Σx_{i} P (X = x_{i})

= (- 2 ) × 0.1 + (- 1 ) × 0.1 + 0 × 0.2 + 1 × 0.2 + 2 × 0.3 + 3 × 0.1 = 0.8

Variance σ^{2} = Σx^{2} P (X = x ) – μ^{2}

= ( – 2 )^{2} × 0.1 + ( – 1)^{2} × 0.1 + 0^{2} × 0.2 + 1^{2} × 0.2 + 2^{2} × 0.3 + 3^{2} × 0.1 – ( 0.8 )^{2} = 2.16.

Question 4.

is the probability distribution of a random variable X. Find the value of k and the variance of X.

Solution:

Question 5.

A random variable X has the following probability distribution.

Find i) k ii) the mean and iii) P(0 < X < 5).

Solution:

Sum of the probability = 1

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

⇒ 10k^{2} + 9k – 1 = 0

⇒ 10k^{2}+ 10k – k – 1 = 0

⇒ 10k(k + 1) -1 (k + 1) = 0

⇒ (10k – 1) (k + 1) = 0

⇒ k = \(\frac{1}{10}\); (k = – 1 is not admissible since probability is non-negative).

II.

Question 1.

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.

Find (i) the value of c

ii) P(X < 1), P (1 < X ≤ 2) and P (0 < X ≤ 3).

Solution:

i) Sum of the probabilities = 1

⇒ 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

⇒ (3c – 1) (c – 1) (c – 2) = 0

⇒ c = \(\frac{1}{3}\) [∵ c ≠ 1, 2].

ii) P(X < 1) = P(X = 0)

= 3c^{3}

= 3 × (\(\frac{1}{27}\)) = \(\frac{1}{9}\)

P(1 < X ≤ 2) = P(X = 2)

= 5c – 1

= 5(\(\frac{1}{27}\)) – 1

= \(\frac{2}{3}\).

P(0 < X ≤ 3) = P(X = 1) + P(X = 2)

= 4c – 10c^{2} + 5c – 1

= – 10c^{2} + 9c – 1

= \(-\frac{10}{9}+\frac{9}{3}-1=\frac{8}{9}\).

Question 2.

The range of a random variable X is {1, 2, 3, …………} and P (X = k) = \(\frac{c^{\mathbf{k}}}{\mathbf{k} !}\). Find the value of c and P (0 < X < 3).

Solution:

∵ Σ P(X = k) = 1