Probability and Random Variable MCQ Quiz - Objective Question with Answer for Probability and Random Variable - Download Free PDF

Concept:

Binomial distribution:

It gives the probability of happening of event 'r' times exactly in 'n' trials.

P(r) = ⁿC_r(p)^r(q)^{n - r}

where n = number of trials, r = number of favourable events, p = probability of happening of an event and q = (1 - p) probability of not happening of an event.

Calculation:

Given:

n = 5 (tossed 5 times), r = 2 (exactly two heads), p = 1/2 (probability of happening event) and q = 1/2 (probability of not happening of an event)

P(r) = nCr(p)r(q)n - r

∴ probability of head occurring exactly 2 times is P(2).

$P(2)=5_{C_2}{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^3\Rightarrow \frac{5}{16}$

Concept:

If X be a random variable with a finite number of outcomes x₁, x_2, x_{3 ....} occurring with probabilities p₁, p₂, p₃....... respectively, the expectation of X is defined as:

$E\left[x\right]=\sum _{i=1}^k{x}_i\cdot p_i=x_1{p}_1+x_2{p}_2\dots$

Application:

It is given that the two sides of a fair coin are labelled as 0 and 1 and the coin is tossed two times independently.

∴ The total possible outcomes can be:

O = {(1,1), (1,0), (0,1), (0,0)}

The random variable X is defined as:

 X = min (M, N)

So,

X₁ = min {(1, 1)} = 1

X₂ = min {(1, 0)} = 0

X₃ = min {(0, 1)} = 0                                                             

X₄ = min {(0, 0)} = 0                     

The probability of occurrence of ‘1’ will be:

$P\left(1\right)=\frac{1}{4}$ 

And the probability of occurrence of ‘0’ will be:

$P\left(0\right)=\frac{3}{4}$ 

We know that:

$E\left[x\right]=\sum _{i=1}^k{x}_i\cdot P_i$

$E\left[x\right]=0\times \frac{3}{4}+1\times \frac{1}{4}$

$E\left[x\right]=0.25$

Concept:

Variance of a random variable ‘y’ is given by:

Var[y] = E[y²] – E²[y]

Properties of mean:

1) E[K] = K, Where K is some constant

2) E[c X] = c. E[X], Where c is some constant

3) E[a X + b] = a E[X] + b, Where a and b are constants

4) E[X + Y] = E[X] + E[Y]

Application:

Variance of y = E[(2x + 3)²] – (E[2x + 3])²

= E[4x² + 12x + 9] – (E[2x + 3])²

= 4E[x²] + 12E[x] + 9 – (E[2x] + E[3])²

= 4E[x2] + 12E[x] + 9 – (2E[x] + 3)2

= 4E[x2] + 12E[x] + 9 – (4E²[x] + 9 + 12E[X])

= 4E[x2] + 12E[x] + 9 – 4E2[x] – 9 – 12E[X]

= 4E[x2] – 4E2[x]

= 4[E[x²] – E²[x]]

This can be written as:

= 4 (variance of x), i.e.

The variance of y = 4 times the variance of x

${\sigma }_y^2=4{\sigma }_x^2$

Properties of Variance:

1) V[K] = 0, Where K is some constant.

2) V[cX] = c2 V[X]

3) V[aX + b] = a2 V[X]

4) V[aX + bY] = a2 V[X] + b2 V[Y] + 2ab Cov(X,Y)

Cov.(X,Y) = E[XY] - E[X].E[Y]

Concept:-

P{X > 1} is representing the probability for all the values of random variable X > 1.

$\Rightarrow P(X>1)={\int }_1^{\infty }{e}^{-x}dx$

Calculation:

Given: f (x) = e-x, 0 < x < ∞

$\Rightarrow P(X>1)={\int }_1^{\infty }{e}^{-x}dx$

$\Rightarrow P(X>1)={\frac{e^{-x}}{-1}|}_1^{\infty }$

$\Rightarrow P(X>1)={\frac{e^{-x}}{-1}|}_1^{\infty }$

⇒ P(X > 1) = - (e^-∞ - e-1)

⇒  P(X > 1) = 1/e

So, the correct answer is option 1

Concept:

The probability density function of a zero-mean Gaussian variable is as shown:

Mathematically, the density function of a Gaussian Random Variable is defined as:

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Given distribution has zero mean i.e. μ = 0, so the above distribution can be written as:

$f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{x^2}{2{\sigma }^2}}$

The Fourier Transform of a Gaussian distribution is as shown:

$F\left(\omega \right)=\sqrt{2{\sigma }^2\pi }{e}^{\frac{2{\sigma }^2{\omega }^2}{4}}$

Clearly, the phase spectrum is constant and with the same standard deviation for the given two random signals, the phase is uniform from -π to +π.

Analysis:

A first-order low pass RC filter is shown below:

The frequency response of the RC filter is:

$\mathrm{H}\left(\omega \right)=\frac{1}{1+\mathrm{j}\omega \mathrm{R}\mathrm{C}}$

The power spectral density is given as:

S_XX(ω) = K

The output power spectral density is related to the input power spectral density by the following relation:

$S_{YY}\left(\omega \right)={\left|H\left(\omega \right)\right|}^2{S}_{XX}\left(\omega \right)$

$S_{YY}\left(\omega \right)=\frac{1}{1+{\left(\omega RC\right)}^2}.K$

$S_{YY}\left(\omega \right)=K.\frac{1}{2RC}.\frac{2\left[1/RC\right]}{{\omega }^2+{\left[1/\left(RC\right)\right]}^2}$

Also, the autocorrelation and the power spectral density form a Fourier transform pair, i.e.

${\mathrm{R}}_{\mathrm{Y}\mathrm{Y}}\left(\omega \right)\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }{\mathrm{S}}_{\mathrm{Y}\mathrm{Y}}\left(\omega \right)$

Thus, using the relation:

${\mathrm{e}}^{-\mathrm{a}\left|\mathrm{t}\right|}\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\frac{2\mathrm{a}}{{\mathrm{a}}^2+{\omega }^2}$ we have inverse Fourier transform of  ${\mathrm{S}}_{\mathrm{Y}\mathrm{Y}}\left(\omega \right)$ as:

${\mathrm{R}}_{\mathrm{Y}\mathrm{Y}}\left(\tau \right)=\mathrm{K}.\frac{1}{2\mathrm{R}\mathrm{C}}.{\mathrm{e}}^{-\frac{\left|\tau \right|}{\mathrm{R}\mathrm{C}}}$

Now, the average power $\mathrm{E}\left[{\mathrm{Y}}^2\left(\mathrm{t}\right)\right]$ will be:

$\mathrm{E}\left[{\mathrm{Y}}^2\left(\mathrm{t}\right)\right]={\mathrm{R}}_{\mathrm{Y}\mathrm{Y}}\left(0\right)=\mathrm{K}.\frac{1}{2\mathrm{R}\mathrm{C}}$

$\therefore \mathrm{E}\left[{\mathrm{Y}}^2\left(\mathrm{t}\right)\right]=\frac{\mathrm{K}}{2\mathrm{R}\mathrm{C}}$

The spectral density of white noise is Uniform and the autocorrelation function of White noise is the Delta function.

Explanation:

The power spectral density is basically the Fourier transform of the autocorrelation function of the power signal, i.e.

$S_x\left(f\right)=F.T.\left\{R_x\left(\tau \right)\right\}$

Also, the inverse Fourier transform of a constant function is a unit impulse.

The power spectral density of white noise is given by:

${\mathrm{S}}_{\mathrm{X}}\left(\mathrm{f}\right)=\frac{\eta }{2}$ for all frequency 'f', i.e.

Now auto-correlation is inverse Fourier transform (IFT) of power spectral density function.

${\mathrm{R}}_{\mathrm{x}}\left(\tau \right)\stackrel{\mathrm{I}\mathrm{F}\mathrm{T}}{\to }{\mathrm{S}}_{\mathrm{X}}\left(\mathrm{f}\right)$

The inverse Fourier transform of the power spectrum of white noise will be an impulse as shown:

${\mathrm{R}}_{\mathrm{x}}\left(\tau \right)=\frac{\eta }{2}\delta \left(\tau \right)$

Concept: ${\mathrm{f}}_{\mathrm{x}\mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)=\{\begin{array}{cc}\left(\mathrm{x}+\mathrm{y}\right),& \begin{array}{cc}0\le \mathrm{x}\le 1,& 0\le \mathrm{y}\le 1\end{array}\\ 0,& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$is given as joint pdf.  The probability of the required region can be found by integrating the joint pdf over the given region. 

Application: We have, ${\mathrm{f}}_{\mathrm{x}\mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)=\{\begin{array}{cc}\left(\mathrm{x}+\mathrm{y}\right),& \begin{array}{cc}0\le \mathrm{x}\le 1,& 0\le \mathrm{y}\le 1\end{array}\\ 0,& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$

Now, region $\mathrm{x}+\mathrm{y}\le 1$ implies

$\mathrm{y}\le 1–\mathrm{x}$

plotting the region in $\mathrm{x}\mathrm{y}$ plane we have

The rectangle bounded by $0\le \mathrm{x}\le 1$ and $0\le \mathrm{y}\le 1$ defined the region for which random variables $\mathrm{x}$ and $\mathrm{y}$ are defined. The shaded region shows the region $\mathrm{x}+\mathrm{y}\le 1$.

Now,

$\begin{array}{l}\Rightarrow \mathrm{P}\left(\mathrm{X}+\mathrm{Y}\le 1\right)=\underset{\mathrm{y}=0}{\overset{1}{\int }}\underset{\mathrm{x}=0}{\overset{1-\mathrm{y}}{\int }}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}\\ =\underset{\mathrm{y}=0}{\overset{1}{\int }}\left(\frac{{\left(1-\mathrm{y}\right)}^2}{2}+\mathrm{y}\left(1-\mathrm{y}\right)\right)\mathrm{d}\mathrm{y}\\ =\frac{1}{2}\underset{\mathrm{y}=0}{\overset{1}{\int }}\left[\left(1-\mathrm{y}\right)\left[1-\mathrm{y}+2\mathrm{y}\right]\right]\mathrm{d}\mathrm{y}\\ =\frac{1}{2}\underset{\mathrm{y}=0}{\overset{1}{\int }}\left(1-{\mathrm{y}}^2\right)\mathrm{d}\mathrm{y}\\ =\frac{1}{2}{\left[\mathrm{y}-\frac{{\mathrm{y}}^3}{3}\right]}_0^1\\ =\frac{1}{2}\left[1-\frac{1}{3}\right]\\ \Rightarrow \mathrm{P}\left(\mathrm{X}+\mathrm{Y}\le 1\right)=\frac{1}{3}\end{array}$

Concept Used:-

An expected value is the “average” value of a random variable. The expected value of a random variable is denoted by E(x) and known as its mean.

The expected value of a random variable is given as,

$E(x)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}xf(x)dx$

Explanation:

Given function is,

\(f(x) = \left\{ \begin{array}{rcl} x-{5\over2}, & 02 \end{array}\right.\)

The expected value or mean of a random variable x for this function with above formula can be given as,

$$\begin{gathered} E(x)={\int }_0^{1}x(x-\frac{5}{2})dx+{\int }_1^{2}x\cdot 2xdx \\ E(x)={\int }_0^1(x^2-\frac{5x}{2})dx+{\int }_1^{2}2x^{2}dx \\ E(x)={\frac{x^3}{3}|}_0^1-{\frac{5x^2}{4}|}_0^1+{\frac{2x^3}{3}|}_1^2 \\ E(x)=\frac{1}{3}-\frac{5}{4}+\frac{2(8-1)}{3} \\ E(x)=\frac{15}{4} \\ E(x)=3.75 \end{gathered}$$

Thus, the mean of a random variable is 3.75.

So, the correct option is 3.

Concept:

E[x] = Mean of Random variable X

X = discrete random variable

The expectation is given by the formula:

$E\left[x\right]=\sum _{i=1}^n{x}_{i}P\left(x_i\right)$ 

x_i = i^th random variable

P(x_i) = Probability of x_i

Calculation:

X: set off positive odd numbers less than 100, i.e.

X: {1, 3, 5, 7, 9, 11, …, 93, 95, 97, 99}

Since X is uniformly chosen, the probability of each random variable will be:

$P\left(x_i\right)=\frac{1}{n}$ 

Where n = total number of random variables

Here, n = number of odd numbers less than 100

n = 50

$P\left(x_i\right)=\frac{1}{50}$ 

$E\left[x\right]=\sum _{i=1}^n{x}_i\left(P\left(x_i\right)\right)=\sum _{i=1}^n{x}_i\left(\frac{1}{50}\right)$ 

$E\left[x\right]=\frac{1}{50}\left[1+3+5+7\dots +93+95+97+99\right]$ 

Clearly an AP is formed with n = 50, a = 1 and l = 99.

The sum of an AP is given by the formula:

$S=\frac{n}{2}\left\{a+l\right\}$ 

$E\left[x_i\right]=\frac{1}{50}\left[\frac{50}{2}\left\{1+99\right\}\right]$ 

$=\frac{1}{2}\times 100$