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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Test  >  Communication System  >  Test: Mutual Information - Electronics and Communication Engineering (ECE) MCQ

Mutual Information - Free MCQ Practice Test with solutions, GATE ECE Engineering


MCQ Practice Test & Solutions: Test: Mutual Information (8 Questions)

You can prepare effectively for Electronics and Communication Engineering (ECE) Communication System with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Mutual Information". These 8 questions have been designed by the experts with the latest curriculum of Electronics and Communication Engineering (ECE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 8

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*Multiple options can be correct
Test: Mutual Information - Question 1
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Read the following expression regarding mutual information I(X;Y). Which of the following expressions is/are correct

Detailed Solution: Question 1

Mutual Information measures the amount of information that one Random variable contains about another Random variable.
The mutual information between two jointly distributed discrete Random variables X and Y is given by:

In terms of Entropy, this is written as:
I(X ; Y) = H(X) – H(X/Y)        ---(1)
(Option (b) is correct)
Also, the conditional entropy states that:
H(X, Y) = H(Y/X) + H(X)
H(X, Y) = H(X/Y) + H(Y)
From above equations we can write:
H(X/Y) = H(X, Y) – H(Y)        ---(2)
Using Equations (1) and (2), we can write:
I(X ; Y) = H(X) – [H(X, Y) – H(Y)]
I(X ; Y) = H(X) + H(Y) – H(X, Y)        ---(3)
(Option (c) is correct)

Test: Mutual Information - Question 2
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Which of the following statements are correct?
(A) A given source will have maximum entropy if the produced are statistically independent
(B) As the bandwidth approaches infinity, the channel capacity becomes zero.
(C) For binary transmission the baud rate is always equal to bit rate
(D) The mutual information of a channel with independent input and output is constant
(E) Nat is a unit of information
Choose the correct answer from the options given below:
(1) (A) and (E) only
(2) (A) and (B) only
(3) (C) and (E) only
(4) (A), (D) and (E) only

Detailed Solution: Question 2

Statement (A):- A given source will have maximum entropy if messages produced are equally probable. So statement A given is wrong.
Statement (B):- From Shannon's channel capacity theorem:-

Statement B is wrong.
Statement (c):-
Band rate = 
For binary transmission M = 2
Band rate  
Band rate = Bitrate
Statement (C) is correct.
Statement (D):- it is False
Mutual information I (X, Y) = H (X) – H (X/Y)
= H (Y) – H (Y/X)
For binary symmetric channel
I (X, Y) is dependent on source
Probabilities it is not constant.
Statement (E):- It is correct.
Nat, bit, or Hartley is unit of Information,
Option (3) is correct.

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*Answer can only contain numeric values
Test: Mutual Information - Question 3
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Let (X1, X2) be independent random varibales. X1 has mean 0 and variance 1, while X2 has mean 1 and variance 4. The mutual information I(X1 ; X2) between X1 and X2 in bits is_______.


Detailed Solution: Question 3

Mutual information of two random variables is a measure to tell how much one random variable tells about the other.
It is mathematically defined as:
I(X1, X2) = H(X1) – H(X1/X2)
Application:
Since X1 and X2 are independent, we can write:
H(X1/X2) = H(X1)
I(X1,X2 ) = H(X1) – H(X1)
= 0

*Answer can only contain numeric values
Test: Mutual Information - Question 4
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For the channel shown below if the source generates two symbols m0 and m1 with a probability of 0.6 and 0.4 respectively. The probability of error if the receiver uses MAP coding will be_______(correct up to two decimal places)


Detailed Solution: Question 4

If r0 is received:
P(m0) P(r0­/m0) = 0.6 × 0.6 = 0.36
Also,
P(m1) P(r0­/m1) = 0.4 × 0 = 0
If r1 is received:
P(m0) P (r1/m0) =  (0.6) (0.3)
= 0.18
P(m1­) P(r1/m1)
= (0.4) (0.7)
= 0.28
If r2 is received:
P(m0) P(r2/m0)
(0.6) (0.1)
= 0.06
P(m1) P(r2/m1)
= (0.4) (0.3)
= 0.12
Probability of correct detection
Pc = 0.12 + 0.28 + 0.36
= 0.76
Pe = 1 – 0.76
= 0.24

*Answer can only contain numeric values
Test: Mutual Information - Question 5
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For the channel shown below, if the source generates M0 and M1 symbols. The probability of error using ML decoding is


Detailed Solution: Question 5

If r0 is received
P (m0) P (r0 / m0) = 0.5 × 0.5 = 0.30
P (m1) P (r0 / m1) = 0.5 × 0 = 0
If r1­ is received
P (m1) P(r1/m0)
(0.5) (0.3)
= 0.15
P (m1) P(r1/m1)
(0.5) (0.7)
= 0.35
If r2 is received
P (m0) P(r2/m0)
= (0.5) (0.1)
= 0.05
P (m1) P(r2/m1)
(0.5) (0.3)
= 0.15
Probability of correct = 0.15 + 0.35 + 0.30
⇒ 0.80
Probability of error = 1 – 0.80
= 0.20 

*Answer can only contain numeric values
Test: Mutual Information - Question 6
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Consider a Binary - channel
P(x1) = 0.5
P(x2) = 0.5

Find the mutual Information in bits/symbol


Detailed Solution: Question 6


P(y1) = 3/8
P(y2) = 5/8
Mutual Information I(xy)
I(xy) = H(x) - H(x/y)
= H(y) - H(y/x)
H(y) - Σ P(yj) log2 P(yi)
= - [0.375 log2(0.375) + 0.625 log2(0.625)]

I(xy) = 0.05 bits/symbol

*Answer can only contain numeric values
Test: Mutual Information - Question 7
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In data communication using error detection code, as soon as an error is detected, an automatic request for retransmission (ARQ) enables retransmission of data. such binary erasure channel can be modeled as shown:

If P = 0.2 and both symbols are generated with equal probability. Then mutual information I(x, y) is _______.


Detailed Solution: Question 7


I (xy) = (1 - p) H (x)

= 0.4 [2 log 2] ( log used is base 2)
⇒ 0.8

*Answer can only contain numeric values
Test: Mutual Information - Question 8
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A binary channel matrix is given by

Given, P(x1) = 1/3 and P(x2) = 2/3. The value of H(Y) is ________bit/symbol.


Detailed Solution: Question 8

The channel matrix can be represented as

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About this Electronics and Communication Engineering (ECE) Practice Test

This test contains 8 MCQs on Mutual Information with detailed solutions for each question. It is part of the Communication System course on EduRev, designed to align with the current Electronics and Communication Engineering (ECE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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