6 Most VSAQ’s of Random Variables and Distribution Chapter in Inter 2nd Year Maths-2A (TS/AP)

Table of Contents

2 Marks

VSAQ-1 : A Poisson variable satisfies P(X=1)=P(X=2), find P(X=5)

We have $$P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}$$ $$\lambda > 0$$

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

$$\Rightarrow \frac{\lambda e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}$$

$$\Rightarrow \frac{\lambda}{1} = \frac{\lambda^2}{2}$$

$$\Rightarrow \lambda^2 = 2\lambda$$

$$\Rightarrow \lambda(\lambda – 2) = 0$$

$$\Rightarrow \lambda = 2$$

$$P(X = 5) = \frac{e^{-2} 2^5}{5!}$$

VSAQ-2 : The mean and variance of a binomial distribution are 4 and 3 respectively. Find the distribution and find P(X≥1)

Given mean $$np = 4$$ variance $$npq = 3$$

Now, $$(np)q = 3$$

$$\Rightarrow (4)q = 3$$

$$\Rightarrow q = \frac{3}{4}$$

$$\Rightarrow p = 1 – q = 1 – \frac{3}{4} = \frac{4-3}{4} = \frac{1}{4}$$

Take $$np = 4$$

$$\Rightarrow n\left(\frac{1}{4}\right) = 4$$

$$\Rightarrow n = 4 \times 4 = 16$$

Binomial distribution is

$$P(X \geq 1) = 1 – P(X=0) = 1 – q^n = 1 – \left(\frac{3}{4}\right)^n = 1 – \left(\frac{3}{4}\right)^{16}$$

VSAQ-3 : For a binomial distribution with mean 6 and variance 2. Find the first two terms of the distribution

Given mean $$np = 6$$ variance $$npq = 2$$

$$(np)q = 2$$

$$\Rightarrow 6(q) = 2$$

$$\Rightarrow q = \frac{2}{6} = \frac{1}{3}$$

$$\Rightarrow p = 1 – q = 1 – \frac{1}{3} = \frac{2}{3}$$

Take $$np = 6$$

$$\Rightarrow n \cdot \frac{2}{3} = 6$$

$$\Rightarrow n = \frac{6}{\frac{2}{3}} = \frac{6 \cdot 3}{2} = 9$$

VSAQ-4 : The probability that a person chosen at random is left-handed (in handwriting) is 0.1. What is the probability that in a group of 10 people, there is one who is left-handed?

Probability of getting a left handed person = probability of success $$p = 0.1 = \frac{1}{10}$$

$$\Rightarrow q = 1 – \frac{1}{10} = \frac{9}{10}$$ Also $$n = 10$$

We know $$p(X=r) = {n}{r} q^{n-r} p^r$$

$$\Rightarrow p(X=1) = 10C1 \cdot \left(\frac{9}{10}\right)^{10-1} \cdot \left(\frac{1}{10}\right)^1 = 10 \cdot \left(\frac{9}{10}\right)^9 \cdot \left(\frac{1}{10}\right) = \left(\frac{9}{10}\right)^9$$

VSAQ-5 : On an average rain falls on 12 days in every 30 days, find the probability that, rain will fall on just 3 days of a given week

Probability of getting rain $$P = \frac{12}{30} = \frac{2}{5}$$

$$q = 1 – P = 1 – \frac{2}{5} = \frac{3}{5}$$

In a week we have $$n = 7$$

$$P(X=3) = {n}{r} q^{n-r} p^r = 7C3 \left(\frac{3}{5}\right)^4 \left(\frac{2}{5}\right)^3 = 35 \cdot \left(\frac{3}{5}\right)^4 \cdot \left(\frac{2}{5}\right)^3 = 35 \times 3^4 \times 2^3 / 5^7$$

VSAQ-6 : If the mean and variance of a binomial variable X are 2.4 & 1.44 respectively, then find P(1<x≤4)

Given Mean = $$np = 2.4$$

Variance $$npq = 1.44$$

Dividing (2) by (1), $$\frac{npq}{np} = \frac{1.44}{2.4} = \frac{3}{5}$$

$$q = \frac{3}{5}$$

$$\Rightarrow p = 1 – q = 1 – \frac{3}{5} = \frac{2}{5}$$

Take $$np = 2.4$$

$$\Rightarrow n\left(\frac{2}{5}\right) = 2.4$$

$$\Rightarrow n = 2.4\left(\frac{5}{2}\right) = 6$$

$$P(1 < X \leq 4) = P(X=2) + P(X=3) + P(X=4)$$

$$= 6C2\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^2 + 6C3\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right)^3 + 6C4\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^4$$

$$= 15\left(\frac{3^4 \cdot 2^2}{5^6}\right) + 20\left(\frac{3^3 \cdot 2^3}{5^6}\right) + 15\left(\frac{3^2 \cdot 2^4}{5^6}\right)$$

$$= 3 \times 5\left(\frac{3^4 \times 2^2}{5^6}\right) + 2 \times 5\left(\frac{3^3 \times 2^3}{5^6}\right) + 3 \times 5\left(\frac{3^2 \times 2^4}{5^6}\right)$$

$$= \frac{36}{15625}(135 + 120 + 60) = \frac{2268}{3125}$$