Random Variables

This chapter introduces the fundamental concept of random variables, transforming experimental outcomes into numerical values for rigorous analysis. Mastery of this topic is crucial for comprehending subsequent advanced probability concepts and is frequently assessed in examinations through problems involving their definitions, distributions, and transformations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Discrete Random Variables |

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We begin with Discrete Random Variables.

Part 1: Discrete Random Variables

This section introduces discrete random variables, their distributions, and fundamental properties essential for analyzing probabilistic systems in computer science. We focus on applying these concepts to solve problems.

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Core Concepts

1. Random Variable (RV)

A random variable is a function that maps outcomes from a sample space to real numbers. We denote random variables by capital letters, e.g., $X$.

Worked Example:

Consider tossing a fair coin three times. Let $X$ be the number of heads. We define the sample space and the value of $X$ for each outcome.

Step 1: Define the sample space $S$.

>

Step 2: Map each outcome to a real number representing the number of heads.

> $X(HHH) = 3$
> $X(HHT) = 2$
> $X(HTH) = 2$
> $X(THH) = 2$
> $X(HTT) = 1$
> $X(THT) = 1$
> $X(TTH) = 1$
> $X(TTT) = 0$

Answer: The set of possible values for $X$ is $\{0, 1, 2, 3\}$.

:::question type="MCQ" question="A fair six-sided die is rolled twice. Let $X$ be the sum of the two rolls. What is the range of $X$?" options=["$\{1, 2, \dots, 6\}$","$\{2, 3, \dots, 12\}$","$\{1, 2, \dots, 12\}$","$\{0, 1, \dots, 6\}$"] answer="$\{2, 3, \dots, 12\}$" hint="Consider the minimum and maximum possible sums." solution="The minimum sum occurs when both rolls are 1, so $1+1=2$. The maximum sum occurs when both rolls are 6, so $6+6=12$. All integer values between 2 and 12 are possible. Therefore, the range of $X$ is $\{2, 3, \dots, 12\}$."
:::

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2. Discrete Random Variable

A random variable is discrete if its range (the set of possible values it can take) is finite or countably infinite.

Worked Example:

Determine if the following random variables are discrete.

Step 1: Define $X_1$ as the number of heads in 10 coin tosses.

> The possible values for $X_1$ are $\{0, 1, 2, \dots, 10\}$. This set is finite. Thus, $X_1$ is a discrete random variable.

Step 2: Define $X_2$ as the time (in minutes) until the first customer arrives at a store.

> The possible values for $X_2$ are any non-negative real number, e.g., $[0, \infty)$. This set is uncountable. Thus, $X_2$ is not a discrete random variable; it is continuous.

Answer: $X_1$ is discrete, $X_2$ is continuous.

:::question type="MCQ" question="Which of the following describes a discrete random variable?" options=["The exact height of a randomly selected student.","The temperature of a room in Celsius.","The number of cars passing a certain point on a road in an hour.","The time it takes for a computer program to execute."] answer="The number of cars passing a certain point on a road in an hour." hint="A discrete random variable takes on a finite or countably infinite number of values." solution="The number of cars is a count, which can only be integer values (0, 1, 2, ...). This is a countably infinite set. Heights, temperatures, and execution times can take on any real value within a range, making them continuous."
:::

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3. Probability Mass Function (PMF)

The PMF of a discrete random variable $X$ gives the probability that $X$ takes on a specific value $x$.

Worked Example:

A biased coin has $P(H) = 0.6$. It is flipped twice. Let $X$ be the number of heads. Find the PMF of $X$.

Step 1: List all possible outcomes and their probabilities.

> $S = \{HH, HT, TH, TT\}$
> $P(HH) = 0.6 \times 0.6 = 0.36$
> $P(HT) = 0.6 \times 0.4 = 0.24$
> $P(TH) = 0.4 \times 0.6 = 0.24$
> $P(TT) = 0.4 \times 0.4 = 0.16$

Step 2: Determine the values $X$ can take and their corresponding probabilities.

> $X=0$: occurs for $TT$. $p_X(0) = P(X=0) = P(TT) = 0.16$
> $X=1$: occurs for $HT, TH$. $p_X(1) = P(X=1) = P(HT) + P(TH) = 0.24 + 0.24 = 0.48$
> $X=2$: occurs for $HH$. $p_X(2) = P(X=2) = P(HH) = 0.36$

Step 3: Verify the PMF properties.

> All $p_X(x) \ge 0$.
> $\sum p_X(x) = 0.16 + 0.48 + 0.36 = 1.00$.

Answer: The PMF is $p_X(0)=0.16$, $p_X(1)=0.48$, $p_X(2)=0.36$.

:::question type="NAT" question="A box contains 3 red balls and 2 blue balls. Two balls are drawn without replacement. Let $X$ be the number of red balls drawn. What is $P(X=1)$?" answer="0.6" hint="Calculate the total number of ways to draw 2 balls and the number of ways to draw 1 red and 1 blue ball." solution="Step 1: Calculate the total number of ways to draw 2 balls from 5.
>

Step 2: Calculate the number of ways to draw 1 red ball from 3 and 1 blue ball from 2.
>
Step 3: Calculate $P(X=1)$.
>
Answer: 0.6"
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4. Cumulative Distribution Function (CDF)

The CDF of a random variable $X$ gives the probability that $X$ takes on a value less than or equal to $x$.

Worked Example:

Using the PMF from the previous example ($p_X(0)=0.16, p_X(1)=0.48, p_X(2)=0.36$), find the CDF $F_X(x)$.

Step 1: Calculate $F_X(x)$ for values less than the smallest possible $X$.

> For $x < 0$, $F_X(x) = P(X \le x) = 0$.

Step 2: Calculate $F_X(x)$ for $x$ between 0 and 1 (inclusive of 0).

> For $0 \le x < 1$, $F_X(x) = P(X \le x) = p_X(0) = 0.16$.

Step 3: Calculate $F_X(x)$ for $x$ between 1 and 2 (inclusive of 1).

> For $1 \le x < 2$, $F_X(x) = P(X \le x) = p_X(0) + p_X(1) = 0.16 + 0.48 = 0.64$.

Step 4: Calculate $F_X(x)$ for $x \ge 2$.

> For $x \ge 2$, $F_X(x) = P(X \le x) = p_X(0) + p_X(1) + p_X(2) = 0.16 + 0.48 + 0.36 = 1$.

Answer: The CDF is:

:::question type="MCQ" question="A discrete random variable $Y$ has the following PMF: $P(Y=1)=0.2$, $P(Y=2)=0.6$, $P(Y=3)=0.2$. What is $F_Y(2.5)$?" options=["0.2","0.6","0.8","1.0"] answer="0.8" hint="The CDF $F_Y(x)$ is the sum of probabilities for all values less than or equal to $x$." solution="Step 1: Recall the definition of CDF.
>

Step 2: Sum the probabilities for $Y$ values less than or equal to 2.5. The possible values of $Y$ that are less than or equal to 2.5 are $Y=1$ and $Y=2$.
>
>
Answer: 0.8"
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5. Expected Value (Mean)

The expected value, or mean, of a discrete random variable $X$ is the weighted average of its possible values, with weights given by their probabilities. It represents the long-run average value of the variable.

Worked Example:

A fair six-sided die is rolled. Let $X$ be the outcome. Calculate $E[X]$.

Step 1: Define the PMF of $X$.

> The possible outcomes are $R_X = \{1, 2, 3, 4, 5, 6\}$.
> For a fair die, $p_X(x) = \frac{1}{6}$ for each $x \in R_X$.

Step 2: Apply the expected value formula.

>

>
>
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Answer: $E[X] = 3.5$.

:::question type="NAT" question="A game involves rolling a fair four-sided die. If the roll is 1, you lose $5. If it's 2 or 3, you win$2. If it's 4, you win $10. Let$ X $be your net winnings. What is$ E[X]$?" answer="2.25" hint="First, determine the PMF of$ X $. Then apply the expected value formula." solution="$Step 1: Determine the possible values of X $ and their probabilities (PMF).
> $X=-5$ if roll is 1. $P(X=-5) = 1/4 = 0.25$
> $X=2$ if roll is 2 or 3. $P(X=2) = 2/4 = 0.50$
> $X=10$ if roll is 4. $P(X=10) = 1/4 = 0.25$
Step 2: Calculate $E[X]$ using the PMF.
>

>
>
Answer: 2.25"
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6. Variance

The variance of a discrete random variable $X$ measures the spread or dispersion of its values around the mean. A higher variance indicates greater variability.

Worked Example:

Using the biased coin example where $X$ is the number of heads in two flips, with PMF $p_X(0)=0.16, p_X(1)=0.48, p_X(2)=0.36$. We previously found $E[X] = 0 \cdot 0.16 + 1 \cdot 0.48 + 2 \cdot 0.36 = 0 + 0.48 + 0.72 = 1.2$. Calculate $\operatorname{Var}(X)$.

Step 1: Calculate $E[X^2]$.

>

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Step 2: Apply the variance formula $\operatorname{Var}(X) = E[X^2] - (E[X])^2$.

> We know $E[X] = 1.2$.
>

>
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Answer: $\operatorname{Var}(X) = 0.48$.

:::question type="MCQ" question="Let $X$ be a discrete random variable with PMF $P(X=1)=0.5$ and $P(X=2)=0.5$. What is $\operatorname{Var}(X)$?" options=["0.125","0.25","0.5","1.0"] answer="0.25" hint="First calculate $E[X]$ and $E[X^2]$." solution="Step 1: Calculate $E[X]$.
>

Step 2: Calculate $E[X^2]$.
>
Step 3: Calculate $\operatorname{Var}(X)$.
>
>
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Answer: 0.25"
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7. Standard Deviation

The standard deviation is the square root of the variance. It provides a measure of spread in the same units as the random variable itself, making it easier to interpret than variance.

Worked Example:

Using the previous example where $\operatorname{Var}(X) = 0.48$, calculate the standard deviation $\sigma_X$.

Step 1: Apply the standard deviation formula.

>

>
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Answer: $\sigma_X \approx 0.6928$.

:::question type="NAT" question="A random variable $X$ has $E[X]=5$ and $E[X^2]=29$. What is the standard deviation of $X$?" answer="2" hint="First calculate the variance using the given values." solution="Step 1: Calculate the variance $\operatorname{Var}(X)$.
>

>
>
Step 2: Calculate the standard deviation $\sigma_X$.
>
>
Answer: 2"
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8. Functions of a Random Variable

If $X$ is a discrete random variable and $g(X)$ is a function of $X$, then $Y = g(X)$ is also a discrete random variable. We can find its PMF, expected value, and variance.

Worked Example:

Let $X$ be the number of heads in two fair coin tosses. The PMF is $p_X(0)=0.25, p_X(1)=0.5, p_X(2)=0.25$. Let $Y = X^2$. Find $E[Y]$.

Step 1: Define the PMF of $X$.

> $p_X(0)=0.25$
> $p_X(1)=0.50$
> $p_X(2)=0.25$

Step 2: Apply the formula for $E[g(X)]$. Here $g(X) = X^2$.

>

>
>
>

Answer: $E[Y] = 1.50$.

:::question type="MCQ" question="A random variable $X$ has PMF $P(X=0)=0.3$, $P(X=1)=0.4$, $P(X=2)=0.3$. Let $Y = 2X+1$. What is $E[Y]$?" options=["1.0","2.0","3.0","4.0"] answer="3.0" hint="Use the linearity of expectation, or first find the PMF of $Y$." solution="Method 1: Using Linearity of Expectation
Step 1: Calculate $E[X]$.
>

Step 2: Apply linearity of expectation for $Y = 2X+1$.
>
>
Method 2: Finding PMF of Y (for verification)
Step 1: Determine the possible values of $Y = 2X+1$ and their probabilities.
> If $X=0$, $Y=2(0)+1=1$. $P(Y=1) = P(X=0) = 0.3$.
> If $X=1$, $Y=2(1)+1=3$. $P(Y=3) = P(X=1) = 0.4$.
> If $X=2$, $Y=2(2)+1=5$. $P(Y=5) = P(X=2) = 0.3$.
Step 2: Calculate $E[Y]$.
>
>
>
Answer: 3.0"
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Common Discrete Distributions

9. Bernoulli Distribution

A Bernoulli random variable models a single trial with two possible outcomes: success (1) or failure (0).

Worked Example:

A component has a probability of $0.9$ of functioning correctly. Let $X=1$ if the component functions correctly and $X=0$ otherwise. Find the variance of $X$.

Step 1: Identify the distribution and its parameter.

> $X \sim \operatorname{Bernoulli}(p)$ with $p=0.9$.

Step 2: Apply the variance formula for Bernoulli distribution.

>

>
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>

Answer: $\operatorname{Var}(X) = 0.09$.

:::question type="NAT" question="A single experiment has a 70% chance of success. Let $X$ be a Bernoulli random variable representing the outcome (1 for success, 0 for failure). What is $E[X]$?" answer="0.7" hint="Recall the mean of a Bernoulli distribution." solution="Step 1: Identify the parameter $p$.
> The probability of success is $p=0.7$.
Step 2: Apply the formula for the expected value of a Bernoulli random variable.
>

>
Answer: 0.7"
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10. Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials.

Worked Example:

A fair coin is flipped 5 times. Let $X$ be the number of heads. Find the probability of getting exactly 3 heads.

Step 1: Identify the distribution and its parameters.

> This is a Binomial distribution with $n=5$ (number of flips) and $p=0.5$ (probability of heads for a fair coin).
> So, $X \sim \operatorname{Binomial}(5, 0.5)$. We want to find $P(X=3)$.

Step 2: Apply the Binomial PMF formula.

>

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Answer: The probability of exactly 3 heads is $0.3125$.

:::question type="MCQ" question="An online server has a 0.8 probability of successfully processing a request. If 10 independent requests are made, what is the expected number of successful requests?" options=["2","4","8","10"] answer="8" hint="Identify the distribution and use its mean formula." solution="Step 1: Identify the distribution and parameters.
> This scenario fits a Binomial distribution, where $n=10$ (number of trials/requests) and $p=0.8$ (probability of success).
> So, $X \sim \operatorname{Binomial}(10, 0.8)$.
Step 2: Use the formula for the mean of a Binomial distribution.
>

>
>
Answer: 8"
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11. Geometric Distribution

The geometric distribution models the number of Bernoulli trials required to get the first success.

Worked Example:

A quality control process involves testing items until a defective one is found. The probability of an item being defective is $0.05$. What is the probability that the first defective item is found on the 5th test?

Step 1: Identify the distribution and its parameter.

> This is a Geometric distribution, as we are looking for the number of trials until the first success (defective item).
> The probability of success (finding a defective item) is $p=0.05$.
> So, $X \sim \operatorname{Geometric}(0.05)$. We want to find $P(X=5)$.

Step 2: Apply the Geometric PMF formula.

>

>
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>

Answer: The probability is approximately $0.0407$.

:::question type="MCQ" question="A biased coin with $P(H)=0.2$ is flipped repeatedly until the first head appears. Let $X$ be the number of flips required. What is the expected number of flips?" options=["2","5","10","20"] answer="5" hint="Identify the distribution and use its mean formula." solution="Step 1: Identify the distribution and parameters.
> This scenario fits a Geometric distribution, where $p=0.2$ (probability of success, i.e., getting a head).
> So, $X \sim \operatorname{Geometric}(0.2)$.
Step 2: Use the formula for the mean of a Geometric distribution.
>

>
>
Answer: 5"
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12. Poisson Distribution

The Poisson distribution models the number of events occurring in a fixed interval of time or space, given a known average rate of occurrence, and events are independent.

Worked Example:

On average, 3 customers arrive at a service desk per hour. Assuming a Poisson process, what is the probability that exactly 2 customers arrive in the next hour?

Step 1: Identify the distribution and its parameter.

> This is a Poisson distribution, as we are counting events (customer arrivals) over a fixed interval (1 hour) with a known average rate.
> The average rate $\lambda = 3$ customers per hour.
> So, $X \sim \operatorname{Poisson}(3)$. We want to find $P(X=2)$.

Step 2: Apply the Poisson PMF formula.

>

>
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Answer: The probability is approximately $0.2241$.

:::question type="NAT" question="A website receives an average of 5 new sign-ups per day. Assuming the number of sign-ups follows a Poisson distribution, what is the variance of the number of sign-ups per day?" answer="5" hint="Recall the variance of a Poisson distribution." solution="Step 1: Identify the parameter $\lambda$.
> The average rate of sign-ups is $\lambda=5$ per day.
Step 2: Apply the formula for the variance of a Poisson random variable.
>

>
Answer: 5"
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Advanced Applications

Worked Example:

An experiment involves $N$ independent trials. Each trial $i$ has a success probability $p_i$. Let $X$ be the total number of successes. Find $E[X]$.

Step 1: Define indicator random variables for each trial.

> Let $X_i$ be an indicator random variable for the $i$-th trial:
>

> For each $X_i$, $E[X_i] = P(X_i=1) = p_i$.

Step 2: Express $X$ as a sum of indicator variables.

> The total number of successes $X$ is the sum of these indicator variables:
>

Step 3: Use linearity of expectation.

>

>
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Answer: $E[X] = \sum_{i=1}^{N} p_i$. This demonstrates the power of linearity of expectation, as the trials do not need to be identically distributed or even independent for this result.

:::question type="MSQ" question="Consider an urn with 5 balls: 3 red and 2 blue. We draw 2 balls without replacement. Let $X$ be the number of red balls drawn. Which of the following statements are true?" options=["The range of $X$ is $\{0, 1, 2\}$.","$P(X=0) = 0.1$.","$E[X] = 1.2$.","This is a Binomial distribution."] answer="The range of $X$ is $\{0, 1, 2\}$,$P(X=0) = 0.1$,$E[X] = 1.2$" hint="Calculate probabilities for each $X$ value. Remember that drawing without replacement implies dependence, ruling out Binomial." solution="Step 1: Determine the range of $X$.
> We draw 2 balls.
> $X=0$: (0R, 2B) $\binom{3}{0}\binom{2}{2} = 1 \times 1 = 1$ way.
> $X=1$: (1R, 1B) $\binom{3}{1}\binom{2}{1} = 3 \times 2 = 6$ ways.
> $X=2$: (2R, 0B) $\binom{3}{2}\binom{2}{0} = 3 \times 1 = 3$ ways.
> Total ways to draw 2 balls from 5: $\binom{5}{2} = 10$ ways.
> The range of $X$ is $\{0, 1, 2\}$. Thus, 'The range of $X$ is $\{0, 1, 2\}$.' is TRUE.

Step 2: Calculate $P(X=0)$.
>

> Thus, '$P(X=0) = 0.1$.' is TRUE.

Step 3: Calculate $E[X]$.
> First, find the full PMF:
> $P(X=0) = 0.1$
> $P(X=1) = \frac{6}{10} = 0.6$
> $P(X=2) = \frac{3}{10} = 0.3$
>

>
> Thus, '$E[X] = 1.2$.' is TRUE.

Step 4: Evaluate 'This is a Binomial distribution.'
> A Binomial distribution assumes independent trials. Drawing without replacement means the probability of drawing a red ball changes after each draw, so the trials are dependent. Therefore, this is NOT a Binomial distribution (it's a Hypergeometric distribution). Thus, 'This is a Binomial distribution.' is FALSE.

Answer: The range of $X$ is $\{0, 1, 2\}$,$P(X=0) = 0.1$,$E[X] = 1.2$"
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="A discrete random variable $X$ has the following PMF: $P(X=1) = 0.4$, $P(X=2) = 0.3$, $P(X=3) = 0.2$, $P(X=4) = 0.1$. Calculate $E[X]$." answer="1.9" hint="Apply the definition of expected value: $\sum x \cdot p_X(x)$." solution="Step 1: Use the formula for expected value.
>

>
>
Answer: 1.9"
:::

:::question type="MCQ" question="Let $X$ be a discrete random variable with PMF given by $P(X=k) = c \cdot k$ for $k=1, 2, 3$, and $P(X=k)=0$ otherwise. What is the value of $c$?" options=["1/3","1/6","1/10","1/12"] answer="1/6" hint="The sum of all probabilities in a PMF must equal 1." solution="Step 1: Use the property that the sum of all probabilities must be 1.
>

>
>
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Answer: 1/6"
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:::question type="MSQ" question="Which of the following statements are true regarding the CDF $F_X(x)$ of a discrete random variable $X$?" options=["$F_X(x)$ is always continuous.","$F_X(x)$ is non-decreasing.","$\lim_{x \to \infty} F_X(x) = 1$.","$P(X=x) = F_X(x) - F_X(x-1)$ for integer $x$ (assuming $F_X(x-1)$ is defined as the limit from the left at $x-1$)."] answer="$F_X(x)$ is non-decreasing,$\lim_{x \to \infty} F_X(x) = 1$,$P(X=x) = F_X(x) - F_X(x-1)$ for integer $x$ (assuming $F_X(x-1)$ is defined as the limit from the left at $x-1$)." hint="Recall the definition and properties of a CDF for discrete random variables. Pay attention to continuity and how PMF relates to CDF." solution="Step 1: Evaluate ' $F_X(x)$ is always continuous.'
> For discrete random variables, the CDF is a step function, which is discontinuous at the values $X$ can take. Thus, this statement is FALSE.

Step 2: Evaluate ' $F_X(x)$ is non-decreasing.'
> This is a fundamental property of any CDF, discrete or continuous. As $x$ increases, the cumulative probability can only stay the same or increase. Thus, this statement is TRUE.

Step 3: Evaluate ' $\lim_{x \to \infty} F_X(x) = 1$.'
> This is another fundamental property of any CDF. As $x$ approaches infinity, it covers all possible outcomes, so the cumulative probability must approach 1. Thus, this statement is TRUE.

Step 4: Evaluate ' $P(X=x) = F_X(x) - F_X(x-1)$ for integer $x$ (assuming $F_X(x-1)$ is defined as the limit from the left at $x-1$).'
> For a discrete random variable taking integer values, $F_X(x) = P(X \le x)$ and $F_X(x-1)$ (more precisely, $\lim_{y \to x^-} F_X(y)$ for step functions) is $P(X \le x-1)$.
> So, $P(X=x) = P(X \le x) - P(X \le x-1) = F_X(x) - F_X(x-1)$. This correctly relates the PMF to the CDF for integer-valued discrete RVs. Thus, this statement is TRUE.

Answer: $F_X(x)$ is non-decreasing,$\lim_{x \to \infty} F_X(x) = 1$,$P(X=x) = F_X(x) - F_X(x-1)$ for integer $x$ (assuming $F_X(x-1)$ is defined as the limit from the left at $x-1$)."
:::

:::question type="NAT" question="A coin is flipped until a head appears. The probability of getting a head on any flip is $p=0.4$. What is the variance of the number of flips required?" answer="3.75" hint="Identify the distribution and use its variance formula." solution="Step 1: Identify the distribution and parameters.
> This scenario describes a Geometric distribution, as we are counting the number of trials until the first success (head).
> The probability of success is $p=0.4$.
> So, $X \sim \operatorname{Geometric}(0.4)$.
Step 2: Use the formula for the variance of a Geometric distribution.
>

>
>
>
Answer: 3.75"
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:::question type="MCQ" question="The number of errors in a software program follows a Poisson distribution with an average of 2 errors per 1000 lines of code. What is the probability of finding exactly 0 errors in a 1000-line segment?" options=["$e^{-2}$","$2e^{-2}$","$e^{-1}$","$2e^{-1}$"] answer="$e^{-2}$" hint="Identify the Poisson parameter $\lambda$ and apply the PMF for $X=0$." solution="Step 1: Identify the Poisson parameter $\lambda$.
> The average number of errors is given as $\lambda=2$ per 1000 lines.
Step 2: Apply the Poisson PMF formula for $X=0$.
>

>
>
>
Answer: $e^{-2}$"
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Summary

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What's Next?

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="Let $X$ be a discrete random variable with the following probability mass function:
$p_X(x) = \begin{cases} c(x+1) & \text{for } x=0, 1, 2 \\ 0 & \text{otherwise} \end{cases}$
What is the value of the constant $c$?" options=["1/6","1/3","1/9","1/4"] answer="1/6" hint="Recall that the sum of all probabilities in a PMF must equal 1." solution="The sum of probabilities must be 1:
$p_X(0) + p_X(1) + p_X(2) = 1$
$c(0+1) + c(1+1) + c(2+1) = 1$
$c(1) + c(2) + c(3) = 1$
$c(1+2+3) = 1$
$6c = 1 \implies c = 1/6$
Thus, the value of $c$ is 1/6."
:::

:::question type="NAT" question="A fair six-sided die is rolled. Let $X$ be the number rolled. Calculate $E[X^2]$." answer="15.166666666666666" hint="First, determine the PMF of $X$. Then use the formula for the expectation of a function of a random variable, $E[g(X)] = \sum_x g(x)p_X(x)$." solution="For a fair six-sided die, the possible values for $X$ are $\{1, 2, 3, 4, 5, 6\}$, and the PMF is $p_X(x) = 1/6$ for each of these values.
We need to calculate $E[X^2]$.
$E[X^2] = \sum_{x=1}^{6} x^2 \cdot p_X(x)$
$E[X^2] = (1^2 \cdot 1/6) + (2^2 \cdot 1/6) + (3^2 \cdot 1/6) + (4^2 \cdot 1/6) + (5^2 \cdot 1/6) + (6^2 \cdot 1/6)$
$E[X^2] = (1/6) \cdot (1 + 4 + 9 + 16 + 25 + 36)$
$E[X^2] = (1/6) \cdot 91$
$E[X^2] = 91/6 \approx 15.166666666666666$"
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:::question type="MCQ" question="In a factory, light bulbs are produced, and the probability that a bulb is defective is $0.05$. A quality control inspector randomly selects bulbs one by one until a defective bulb is found. What is the most appropriate distribution to model the number of bulbs inspected until the first defective one is found?" options=["Binomial Distribution","Poisson Distribution","Geometric Distribution","Bernoulli Distribution"] answer="Geometric Distribution" hint="Consider what each distribution models. Which one describes the number of trials until the first success in a sequence of independent Bernoulli trials?" solution="The scenario describes a sequence of independent Bernoulli trials (each bulb is either defective or not), where we are interested in the number of trials until the first 'success' (finding a defective bulb). This is the definition of a Geometric Distribution.

• Bernoulli Distribution: Models a single trial with two outcomes (success/failure).


• Binomial Distribution: Models the number of successes in a fixed number of trials.


• Poisson Distribution: Models the number of events occurring in a fixed interval of time or space.


• Geometric Distribution: Models the number of trials needed to get the first success."

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:::question type="NAT" question="Let $X$ be a discrete random variable with $E[X] = 3$ and $\operatorname{Var}(X) = 2$. Calculate $\operatorname{Var}(5-2X)$." answer="8" hint="Recall the properties of variance for linear transformations: $\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X)$." solution="We are given $E[X] = 3$ and $\operatorname{Var}(X) = 2$.
We need to calculate $\operatorname{Var}(5-2X)$.
Using the property $\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X)$, where $a=-2$ and $b=5$:
$\operatorname{Var}(5-2X) = (-2)^2 \operatorname{Var}(X)$
$\operatorname{Var}(5-2X) = 4 \cdot \operatorname{Var}(X)$
Substitute the given value for $\operatorname{Var}(X)$:
$\operatorname{Var}(5-2X) = 4 \cdot 2$
$\operatorname{Var}(5-2X) = 8$
The expected value $E[X]$ is not needed for this calculation."
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