Information Theory And Coding Quiz!

1) What is the probability of getting a sum of 9 from two throws of a dice?

1/6

1/8

1/9

1

The probability of getting a sum of 9 from two throws of a dice can be calculated by finding the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there is only one favorable outcome, which is when both dice show a value of 4. The total number of possible outcomes is 6*6 = 36, as each dice has 6 possible outcomes. Therefore, the probability is 1/36.

Explanation

The probability of getting a sum of 9 from two throws of a dice can be calculated by finding the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there is only one favorable outcome, which is when both dice show a value of 4. The total number of possible outcomes is 6*6 = 36, as each dice has 6 possible outcomes. Therefore, the probability is 1/36.

2) If the channel is band-limited to 6 kHz & signal to noise ratio is 16, what would be the capacity of the channel?

15.15 kbps

24.74 kbps

30.12 kbps

52.18 kbps

The capacity of a channel is determined by its bandwidth and the signal-to-noise ratio. In this case, the channel is band-limited to 6 kHz and the signal-to-noise ratio is 16. The formula to calculate the channel capacity is given by: C = B * log2(1 + SNR), where C is the capacity, B is the bandwidth, and SNR is the signal-to-noise ratio. Plugging in the values, we get C = 6 * log2(1 + 16) = 6 * log2(17) = 24.74 kbps. Therefore, the correct answer is 24.74 kbps.

Explanation

The capacity of a channel is determined by its bandwidth and the signal-to-noise ratio. In this case, the channel is band-limited to 6 kHz and the signal-to-noise ratio is 16. The formula to calculate the channel capacity is given by: C = B * log2(1 + SNR), where C is the capacity, B is the bandwidth, and SNR is the signal-to-noise ratio. Plugging in the values, we get C = 6 * log2(1 + 16) = 6 * log2(17) = 24.74 kbps. Therefore, the correct answer is 24.74 kbps.

3) In digital communication system, smaller the code rate, _________are the redundant bits.

Less

More

Equal

Unpredictable

In a digital communication system, the code rate refers to the ratio of useful information bits to the total number of transmitted bits. A smaller code rate means that a larger portion of the transmitted bits are redundant, which are used for error detection and correction. Therefore, the correct answer is "more" as a smaller code rate implies more redundant bits.

Explanation

In a digital communication system, the code rate refers to the ratio of useful information bits to the total number of transmitted bits. A smaller code rate means that a larger portion of the transmitted bits are redundant, which are used for error detection and correction. Therefore, the correct answer is "more" as a smaller code rate implies more redundant bits.

4) Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

1/2

2/5

8/15

9/20

The probability of drawing a ticket with a number that is a multiple of 3 or 5 can be calculated by finding the total number of tickets that are multiples of 3 or 5 and dividing it by the total number of tickets.

Out of the numbers 1 to 20, there are 6 numbers that are multiples of 3 (3, 6, 9, 12, 15, 18) and 4 numbers that are multiples of 5 (5, 10, 15, 20). However, the number 15 is counted twice because it is a multiple of both 3 and 5.

Therefore, the total number of tickets that are multiples of 3 or 5 is 6 + 4 - 1 = 9.

Since there are 20 tickets in total, the probability is 9/20.

Explanation

The probability of drawing a ticket with a number that is a multiple of 3 or 5 can be calculated by finding the total number of tickets that are multiples of 3 or 5 and dividing it by the total number of tickets.

Out of the numbers 1 to 20, there are 6 numbers that are multiples of 3 (3, 6, 9, 12, 15, 18) and 4 numbers that are multiples of 5 (5, 10, 15, 20). However, the number 15 is counted twice because it is a multiple of both 3 and 5.

Therefore, the total number of tickets that are multiples of 3 or 5 is 6 + 4 - 1 = 9.

Since there are 20 tickets in total, the probability is 9/20.

5) A random variable is

A hypothetical list of the possible outcomes of a random phenomenon.

Any phenomenon in which outcomes are equally likely.

Any number that changes in a predictable way in the long run.

A variable whose value is a numerical outcome associated with a random phenomenon.

The correct answer is "a variable whose value is a numerical outcome associated with a random phenomenon." This is the most accurate definition of a random variable as it emphasizes that it is a variable that represents numerical outcomes of a random event or experiment. Random variables can take on different values based on the outcomes of the random phenomenon, and they can be used to model and analyze various probabilistic scenarios.

Explanation

The correct answer is "a variable whose value is a numerical outcome associated with a random phenomenon." This is the most accurate definition of a random variable as it emphasizes that it is a variable that represents numerical outcomes of a random event or experiment. Random variables can take on different values based on the outcomes of the random phenomenon, and they can be used to model and analyze various probabilistic scenarios.

6) Susan took two tests. The probability of her passing both tests is 0.6. The probability of her passing the first test is 0.8. What is the probability of her passing the second test given that she has passed the first test?

0.75

0.65

0.80

0.70

The probability of Susan passing both tests is 0.6. The probability of her passing the first test is 0.8. To find the probability of her passing the second test given that she has passed the first test, we can use the formula for conditional probability. The formula states that the probability of event B given event A is equal to the probability of both events A and B happening divided by the probability of event A. In this case, the probability of passing both tests (A and B) is 0.6, and the probability of passing the first test (A) is 0.8. So, the probability of passing the second test (B) given that she has passed the first test is 0.6/0.8 = 0.75.

Explanation

The probability of Susan passing both tests is 0.6. The probability of her passing the first test is 0.8. To find the probability of her passing the second test given that she has passed the first test, we can use the formula for conditional probability. The formula states that the probability of event B given event A is equal to the probability of both events A and B happening divided by the probability of event A. In this case, the probability of passing both tests (A and B) is 0.6, and the probability of passing the first test (A) is 0.8. So, the probability of passing the second test (B) given that she has passed the first test is 0.6/0.8 = 0.75.

7) In a recent survey in a Statistics class, it was determined that only 60% of the students attend class on Thursday. From past data, it was noted that 98% of those who went to class on Thursday pass the course, while only 20% of those who did not go to class on Thursday passed the course.

What percentage of students is expected to pass the course?

0.66

0.24

0.45

0.52

Based on the given information, 60% of the students attend class on Thursday. Out of those who attend class on Thursday, 98% pass the course. Therefore, the percentage of students who attend class on Thursday and pass the course is 0.60 * 0.98 = 0.588. On the other hand, 40% of the students do not attend class on Thursday. Out of those who do not attend class on Thursday, only 20% pass the course. Therefore, the percentage of students who do not attend class on Thursday and pass the course is 0.40 * 0.20 = 0.08. To find the overall percentage of students expected to pass the course, we add these two percentages: 0.588 + 0.08 = 0.668. Hence, the expected percentage of students to pass the course is 0.668, which is approximately 0.66.

Explanation

Based on the given information, 60% of the students attend class on Thursday. Out of those who attend class on Thursday, 98% pass the course. Therefore, the percentage of students who attend class on Thursday and pass the course is 0.60 * 0.98 = 0.588. On the other hand, 40% of the students do not attend class on Thursday. Out of those who do not attend class on Thursday, only 20% pass the course. Therefore, the percentage of students who do not attend class on Thursday and pass the course is 0.40 * 0.20 = 0.08. To find the overall percentage of students expected to pass the course, we add these two percentages: 0.588 + 0.08 = 0.668. Hence, the expected percentage of students to pass the course is 0.668, which is approximately 0.66.

8) The weight of written reports produced in a certain department has a Normal distribution with a mean of 60 g and a standard deviation of 12 g. The probability that the next report will weigh less than 45 g is

.8944

.1042

.3944

.1056

The given question states that the weight of written reports in a certain department follows a Normal distribution with a mean of 60 g and a standard deviation of 12 g. To find the probability that the next report will weigh less than 45 g, we need to calculate the z-score first. The z-score is calculated by subtracting the mean from the given value (45 g) and dividing it by the standard deviation (12 g). In this case, the z-score is -1.25. Using a standard Normal distribution table or a calculator, we can find that the probability of a z-score less than -1.25 is approximately 0.1056. Therefore, the correct answer is .1056.

Explanation

The given question states that the weight of written reports in a certain department follows a Normal distribution with a mean of 60 g and a standard deviation of 12 g. To find the probability that the next report will weigh less than 45 g, we need to calculate the z-score first. The z-score is calculated by subtracting the mean from the given value (45 g) and dividing it by the standard deviation (12 g). In this case, the z-score is -1.25. Using a standard Normal distribution table or a calculator, we can find that the probability of a z-score less than -1.25 is approximately 0.1056. Therefore, the correct answer is .1056.

9) A bag contains 2 red, 3 green, and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

10/21

11/21

2/7

5/7

The probability of not drawing a blue ball can be calculated by finding the probability of drawing a non-blue ball for each draw. On the first draw, there are 7 balls remaining, out of which 5 are non-blue. Therefore, the probability of not drawing a blue ball on the first draw is 5/7. On the second draw, there are 6 balls remaining, out of which 4 are non-blue. Therefore, the probability of not drawing a blue ball on the second draw is 4/6. The overall probability of not drawing a blue ball is the product of these probabilities, which is (5/7) * (4/6) = 20/42 = 10/21.

Explanation

The probability of not drawing a blue ball can be calculated by finding the probability of drawing a non-blue ball for each draw. On the first draw, there are 7 balls remaining, out of which 5 are non-blue. Therefore, the probability of not drawing a blue ball on the first draw is 5/7. On the second draw, there are 6 balls remaining, out of which 4 are non-blue. Therefore, the probability of not drawing a blue ball on the second draw is 4/6. The overall probability of not drawing a blue ball is the product of these probabilities, which is (5/7) * (4/6) = 20/42 = 10/21.

10) In a recent survey in a Statistics class, it was determined that only 60% of the students attend class on Thursday. From past data it was noted that 98% of those who went to class on Thursday pass the course, while only 20% of those who did not go to class on Thursday passed the course.

Given that a student passes the course, what is the probability that he/she attended classes on Thursday.

0.61

0.85

0.93

0.55

The probability that a student attended class on Thursday given that they passed the course can be calculated using Bayes' theorem. The theorem states that the probability of A given B is equal to the probability of B given A multiplied by the probability of A, divided by the probability of B. In this case, A represents attending class on Thursday and B represents passing the course. The probability of attending class on Thursday given that the student passed the course is 0.85, which means that there is an 85% chance that a student who passed the course attended class on Thursday.

Explanation

The probability that a student attended class on Thursday given that they passed the course can be calculated using Bayes' theorem. The theorem states that the probability of A given B is equal to the probability of B given A multiplied by the probability of A, divided by the probability of B. In this case, A represents attending class on Thursday and B represents passing the course. The probability of attending class on Thursday given that the student passed the course is 0.85, which means that there is an 85% chance that a student who passed the course attended class on Thursday.

11) In class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

21/46

25/117

1/50

3/25

The probability of selecting 1 girl and 2 boys can be calculated by finding the probability of selecting 1 girl from the 10 available girls (10/25) and then multiplying it by the probability of selecting 2 boys from the 15 available boys (15/24 * 14/23). Simplifying this expression gives us (10/25) * (15/24) * (14/23) = 21/46. Therefore, the correct answer is 21/46.

Explanation

The probability of selecting 1 girl and 2 boys can be calculated by finding the probability of selecting 1 girl from the 10 available girls (10/25) and then multiplying it by the probability of selecting 2 boys from the 15 available boys (15/24 * 14/23). Simplifying this expression gives us (10/25) * (15/24) * (14/23) = 21/46. Therefore, the correct answer is 21/46.

12) In a particular game, a fair die is tossed. If the number of spots showing is either 4 or 5 you win $1, if the number of spots showing is 6 you win $4, and if the number of spots showing is 1, 2, or 3 you win nothing. Let X be the amount that you win. Which of the following is the expected value of X?

$1.00

$2.50

$4.00

$6.00

The expected value of X is calculated by multiplying each possible outcome by its probability and summing them up. In this case, the probability of getting 4 or 5 is 2/6, the probability of getting 6 is 1/6, and the probability of getting 1, 2, or 3 is 3/6. The amount won for 4 or 5 is $1, the amount won for 6 is $4, and the amount won for 1, 2, or 3 is $0. Calculating the expected value, (2/6 * $1) + (1/6 * $4) + (3/6 * $0) = $1.00. Therefore, the expected value of X is $1.00.

Explanation

The expected value of X is calculated by multiplying each possible outcome by its probability and summing them up. In this case, the probability of getting 4 or 5 is 2/6, the probability of getting 6 is 1/6, and the probability of getting 1, 2, or 3 is 3/6. The amount won for 4 or 5 is $1, the amount won for 6 is $4, and the amount won for 1, 2, or 3 is $0. Calculating the expected value, (2/6 * $1) + (1/6 * $4) + (3/6 * $0) = $1.00. Therefore, the expected value of X is $1.00.

13) A man is known to speak the truth 2 out of 3 times. He throws a die and reports that the number obtained is a four. Find the probability that the number obtained is actually a four.

1/6

3/5

2/7

5/7

The probability that the man speaks the truth is 2/3, and the probability of getting a four on a fair die is 1/6. Therefore, the probability that the man speaks the truth and the number obtained is actually a four is (2/3) * (1/6) = 1/9. The probability that the man reports a four, regardless of whether it is true or not, is (2/3) * (1/6) + (1/3) * (5/6) = 2/9 + 5/18 = 9/18 = 1/2. Therefore, the probability that the number obtained is actually a four, given that the man reports a four, is (1/9) / (1/2) = 2/7.

Explanation

The probability that the man speaks the truth is 2/3, and the probability of getting a four on a fair die is 1/6. Therefore, the probability that the man speaks the truth and the number obtained is actually a four is (2/3) * (1/6) = 1/9. The probability that the man reports a four, regardless of whether it is true or not, is (2/3) * (1/6) + (1/3) * (5/6) = 2/9 + 5/18 = 9/18 = 1/2. Therefore, the probability that the number obtained is actually a four, given that the man reports a four, is (1/9) / (1/2) = 2/7.

14) Which of the following is not a random variable?

The heights of randomly-selected buildings in New York City.

The suit of a card randomly-selected from a 52-card deck.

The number of children in randomly-selected households in the United States.

The amount of money won (or lost) by the next person to walk out of a casino in Las Vegas.

The suit of a card randomly-selected from a 52-card deck is not a random variable because it is a discrete and finite set of possibilities (clubs, diamonds, hearts, spades), rather than a numerical value. Random variables are typically defined as numerical quantities that can take on different values based on the outcome of a random event or experiment. In this case, the suit of a card is not a numerical value and therefore does not meet the criteria of a random variable.

Explanation

The suit of a card randomly-selected from a 52-card deck is not a random variable because it is a discrete and finite set of possibilities (clubs, diamonds, hearts, spades), rather than a numerical value. Random variables are typically defined as numerical quantities that can take on different values based on the outcome of a random event or experiment. In this case, the suit of a card is not a numerical value and therefore does not meet the criteria of a random variable.

15) For a baseband system with transmission rate 'r_s' symbols/sec, what would be the required bandwidth?

Rs / 2 Hz

Rs / 4 Hz

Rs / 8 Hz

Rs / 16 Hz

In a baseband system, the required bandwidth is equal to the transmission rate divided by 4 Hz. This is because in baseband systems, the signal occupies a frequency range from 0 Hz to the highest frequency component, which is equal to half the transmission rate. Therefore, to accommodate this signal, the required bandwidth is equal to half the transmission rate divided by 2, which simplifies to rs / 4 Hz.

Explanation

In a baseband system, the required bandwidth is equal to the transmission rate divided by 4 Hz. This is because in baseband systems, the signal occupies a frequency range from 0 Hz to the highest frequency component, which is equal to half the transmission rate. Therefore, to accommodate this signal, the required bandwidth is equal to half the transmission rate divided by 2, which simplifies to rs / 4 Hz.