Virtual Labs

Consider the following question on the binomial distribution. In an experiment,

10

lights are independently and randomly switched on, each with a probability

p \in (0,1)

. Which of the options best describes the probability of observing exactly

4

lights on after this random experiment? (note that

\binom{n}{k}

refers to the binomial coefficient.)

\binom{10}{4}p^6(1-p)^4

Explanation

\binom{10}{4}p^4(1-p)^6

Explanation

p^4

Explanation

p^4(1-p)^6

Explanation

e: None Explanation

Explanation

Consider two random variables

X

and

Y

, taking values in sets

\mathcal{X}=\{0,a\}

and

\mathcal{Y}=\{b,c,5\}

respectively, with the following conditional probabilities,

p_{Y|X}(b|a)=0.2, \hspace{0.2cm} p_{Y|X}(c|a)=0.4, \hspace{0.2cm} p_{Y|X}(c|0)=0.6, \hspace{0.2cm} p_{Y|X}(5|0)=0.1.

What is the value of

p_Y(5)

, if

p_X(0)=0.3

a: 0.62 Explanation

Explanation

b: 0.31 Explanation

Explanation

c: 0.05 Explanation

Explanation

d: 0.28 Explanation

Explanation

Which of the following represents the distribution of a Bernoulli random variable

X

p_X(k)=\binom{n}{k}p^k(1-p)^k, \hspace{0.2cm} \forall k\in\{0,1,...,n\}

Explanation

p_X(0)=0.2,\hspace{0.2cm} p_X(1)=0.1,\hspace{0.2cm} p_X(2)=0.7

Explanation

p_X(0)=0.28,\hspace{0.2cm} p_X(1)=0.72

. Explanation

Explanation

p_X(0)=0.3,\hspace{0.2cm} p_X(1)=0.6

. Explanation

Explanation

Which of the following represents the distribution of a Gaussian random variable

X

with mean

5

and variance

5

\frac{1}{\sqrt{5}}e^{-\frac{(x-25)^2}{10}}

Explanation

\frac{1}{\sqrt{10\pi}}e^{-\frac{(x-5)^2}{50}}

Explanation

\frac{1}{\sqrt{10\pi}}e^{-\frac{(x-5)^2}{10}}

Explanation

\frac{1}{\sqrt{10\pi}}e^{-\frac{|x-5|}{10}}

Explanation

Consider a

4

-length Gaussian random vector

\boldsymbol{X}=(X_1,X_2,X_3,X_4)

, with each component generated independently and identically (i.e.,

X_i

s are i.i.d), from a Gaussian distribution with mean

2

and variance

4

. Select the expression that describes the probability density function of

\boldsymbol{X}

\frac{1}{(8\pi)^2}e^{-\frac{\sum_{i=1}^4(x_i-2)^2}{8}}

Explanation

\frac{1}{(\sqrt{8\pi})}e^{-\frac{(x-2)^2}{8}}

Explanation

\frac{1}{(8\pi)^4}e^{-\frac{\sum_{i=1}^4(x_i-2)^2}{8}}

Explanation

\frac{1}{(4\pi)^2}e^{-\frac{\sum_{i=1}^4(x_i-8)^2}{4}}

. Explanation

Explanation

Binary-input Memoryless Channels