Problem 12 - Applied Statistics and Probability for Engineers

Question

In the transmission of digital information, the probability that a bit has high, moderate, or low distortion is 0.01 , \(0.04,\) and \(0.95,\) respectively. Suppose that three bits are transmitted and that the amount of distortion of each bit is assumed to be independent. Let \(X\) and \(Y\) denote the number of bits with high and moderate distortion of the three transmitted, respectively. Determine the following: (a) Probability that two bits have high distortion and one has moderate distortion (b) Probability that all three bits have low distortion (c) Probability distribution, mean, and variance of \(X\) (d) Conditional probability distribution, conditional mean, and conditional variance of \(X\) given that \(Y=2\)

Step-by-Step Solution

Answer

(a) 0.000012, (b) 0.857375, (c) Mean = 0.03, Variance = 0.0297, (d) Conditional mean = 0, Conditional variance = 0.

Step 1: Understand the Problem

We are dealing with a scenario where each of the transmitted bits can either have high, moderate, or low distortion. The probability for each type of distortion is 0.01 for high, 0.04 for moderate, and 0.95 for low distortion. We are asked to calculate probabilities and distributions related to the number of bits with high and moderate distortion denoted by random variables \(X\) and \(Y\), respectively.

Step 2: Define Probability Events

Define events for the distortion types: High distortion (\(H\)), Moderate distortion (\(M\)), Low distortion (\(L\)). Let the probability of these events be \(P(H) = 0.01\), \(P(M) = 0.04\), \(P(L) = 0.95\). Assume each bit's distortion is independent of the others.

Step 3: Probability of Two High and One Moderate Distortion (a)

The problem asks for the probability that two bits have high distortion (\(H\)) and one has moderate distortion (\(M\)). Use the formula for permutations and probabilities: \[P(X = 2, Y = 1) = \binom{3}{2} \cdot P(H)^2 \cdot P(M) = 3 \cdot (0.01)^2 \cdot 0.04\]Calculate: \[P(X = 2, Y = 1) = 3 \cdot 0.0001 \cdot 0.04 = 1.2 \times 10^{-5}\]

Step 4: Probability of All Low Distortion (b)

Calculate the probability that all three bits have low distortion:\[P(L, L, L) = P(L)^3 = 0.95^3\]Calculate:\[P(L, L, L) = 0.857375\]

Step 5: Probability Distribution of X (c)

Random variable \(X\) (number of bits with high distortion) follows a binomial distribution with parameters \(n = 3\) and \(p = 0.01\). The distribution is:- \(P(X = 0) = \binom{3}{0}(0.01)^0(0.99)^3 = 0.9703\)- \(P(X = 1) = \binom{3}{1}(0.01)^1(0.99)^2 = 0.029403\)- \(P(X = 2) = \binom{3}{2}(0.01)^2(0.99)^1 = 0.000297\)- \(P(X = 3) = \binom{3}{3}(0.01)^3(0.99)^0 = 0.000001\)

Step 6: Mean and Variance of X (c)

For a binomial distribution, the mean \(\mu_X\) is \(np\) and the variance \(\sigma_X^2\) is \(np(1-p)\). So:- Mean: \(\mu_X = 3 \times 0.01 = 0.03\)- Variance: \(\sigma_X^2 = 3 \times 0.01 \times 0.99 = 0.0297\)

Step 7: Conditional Distribution Given Y=2 (d)

Given \(Y = 2\), meaning that two bits have moderate distortion, find \(X\) given this condition. Since two out of three bits are moderate, the only possible value for \(X\) is \(0\) (since no high bits can remain):\[P(X=0|Y=2) = 1\]The conditional mean and variance, therefore:- Conditional mean: \(\mu_{X|Y=2} = 0\)- Conditional variance: \(\sigma_{X|Y=2}^2 = 0\) because variance of a constant is zero.

Key Concepts

Probability distribution

In probability theory, a probability distribution describes how the probabilities are distributed over the outcomes of a random variable. In our exercise, we are examining the distortion of bits during digital information transmission. Each bit can independently have high, moderate, or low distortion, with respective probabilities of 0.01, 0.04, and 0.95.

For example, to understand the distribution of a random variable like the number of bits with high distortion, we employ a probability distribution. This will tell us how likely each number of high-distortion bits is when we send three bits. The probability distribution can be expressed using mathematical functions or tables, and these probabilities must add up to 1.

• Important Concept: The sum of probabilities in a probability distribution across all possible outcomes should equal 1. This ensures that all potential events are accounted for in the model.

Understanding probability distributions helps us make predictions and informed decisions by knowing what outcomes are more likely under given conditions.

Binomial distribution

The binomial distribution is a specific type of probability distribution that applies when there are a fixed number of independent trials, each with two possible outcomes, like "success" and "failure". In our context, success might refer to a bit experiencing high distortion.

In the given problem, we are said to transmit three bits, each possibly experiencing high distortion with a probability of 0.01. This situation is modeled appropriately using a binomial distribution, where:

• The number of trials ( n ) is 3.
• The probability of success ( p ) is 0.01.

We compute probabilities for different numbers of successes using the binomial probability formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]

• Where:
  • \(k\) is the number of successes.
  • \(\binom{n}{k}\) represents the number of combinations.

The mean and variance are found directly using:

• Mean ( \(\mu\) ): \(np\)
• Variance ( \(\sigma^2\) ): \(np(1-p)\)

This distribution is crucial for managing binary outcomes and understanding scenarios like those in the transmission example.

Conditional probability

Conditional probability is the probability of an event occurring given that another event has already occurred. It is especially useful for understanding how information or outcomes can influence each other, which is key in decision-making processes and data analysis.

In the problem at hand, we are asked to find the probability distribution of the number of high distortion bits (denoted by X), given that two bits have moderate distortion (denoted by Y = 2). The formula for conditional probability is:\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]

• Where:
  • \(A\) is the event of interest (e.g., number of high distortion bits).
  • \(B\) is the given condition (e.g., Y = 2).

When dealing with conditional probability, understanding that the context changes based on existing information is vital. This means that in our example, since there are already two bits with moderate distortion, the only possible outcome for X must adjust according to this prior knowledge, leading to the result that X can only equal 0 given Y=2. This eliminates any variability, thus making the conditional variance 0. Such deductions highlight how conditional probability can refine our predictions or expectations in response to new data.