Random Variables And Probability Distributions (VSAQs)

Maths-2A | 10. Random Variables And Probability Distributions – VSAQs:
Welcome to VSAQs in Chapter 10: Random Variables And Probability Distributions. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

Overview:

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 $$np = 2.4$$

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

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

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

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

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

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

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