Discrete Random Variables

A random variable, denoted \(X\), is a function that assigns a numerical value to each outcome \(\omega\) in a random experiment. We write this value as \(X(\omega)\).The possible values of \(X\) are the real numbers that \(X\) can take.

Let \(X\) be the number of heads when tossing 2 fair coins: (red coin) and (blue coin). Find \(X(\textcolor{colordef}{H},\textcolor{colorprop}{T})\).

A random variable is discrete if its set of possible values is finite or countably infinite. This means we can list all possible values.

For a random variable \(X\):

• \((X = x)\): The set of outcomes where \(X\) takes the value \(x\).
• \((X \leq x)\): The set of outcomes where \(X\) is less than or equal to \(x\).
• \((X \geq x)\): The set of outcomes where \(X\) is greater than or equal to \(x\).

Let \(X\) be the number of heads when tossing 2 coins: and . List the outcomes for \((X = 0)\), \((X = 1)\), \((X = 2)\), \((X \leq 1)\), and \((X \geq 1)\).

The probability distribution of a random variable \(X\) lists the probability \(P(X = x_i)\) for each possible value \(x_1,x_2,\dots,x_n\). It can be shown as a table or formula.

For a random variable \(X\) with possible values \(x_1,x_2,\dots,x_n\), we have

• \(0 \leq P(X=x_i) \leq 1\) for all \(i=1,\dots,n\),
• \(\displaystyle\sum_{i=1}^n P(X=x_i) =P(X=x_1)+P(X=x_2)+\dots+P(X=x_n)= 1 \).

Let \(X\) be the number of heads when tossing 2 fair coins: and .

1. List the possible values of \(X\).
2. Find the probability distribution.
3. Create the probability table.
4. Draw the probability distribution graph.

Usually, defining a random variable begins by establishing:

1. a sample space, that is, the set of all possible outcomes,
2. a probability associated with this sample space,
3. a function \(X\) that assigns a number to each outcome in the sample space.

This is quite a lengthy task. However, often, we prefer to directly define a random variable \(X\) with a given probability distribution, relying on the context of the situation being studied. For example, imagine we survey a class of 30 students about their siblings and obtain these results: 10 students have 0 siblings, 12 have 1 sibling, 5 have 2 siblings, and 3 have 3 siblings. We can then define the random variable \(X\) as the number of siblings of a randomly chosen student, with this probability distribution:

\(x\)	0	1	2	3
\(P(X = x)\)	\(\frac{10}{30}\)	\(\frac{12}{30}\)	\(\frac{5}{30}\)	\(\frac{3}{30}\)

The theorem below shows that it is always possible to construct a sample space, a probability, and a function \(X\) to obtain a random variable with this probability distribution.

Suppose you have possible values \(x_1, x_2, \ldots, x_n\) and probabilities \(p_1, p_2, \ldots, p_n\).
If:

• \(0 \leq p_i \leq 1\) for each \(i = 1, 2, \ldots, n\),
• \(\displaystyle\sum_{i=1}^n p_i = p_1 + p_2 + \cdots + p_n = 1\),

then there exists a random variable \(X\) with the probability distribution \(P(X = x_i) = p_i\) for each \(i = 1, 2, \ldots, n\).

In practice, we often define a random variable \(X\) directly by specifying its probability distribution. The key is to ensure that this distribution is valid, meaning it satisfies the conditions for a probability distribution: all probabilities must be non-negative and sum to 1.

We survey a class of 30 students about their siblings and obtain these results: 10 students have 0 siblings, 12 have 1 sibling, 5 have 2 siblings, and 3 have 3 siblings. We define a random variable \(X\) as the number of siblings of a randomly chosen student, with this probability distribution:

\(x\)	0	1	2	3
\(P(X = x)\)	\(\frac{10}{30}\)	\(\frac{12}{30}\)	\(\frac{5}{30}\)	\(\frac{3}{30}\)

Determine if this probability distribution is valid.

The expected value of a random variable \(X\) is the "average you’d expect if you repeated the experiment many times". It’s found by taking all possible values, multiplying each by its probability, and adding them up — essentially a weighted average where the probabilities act as the weights.

For a random variable \(X\) with possible values \(x_1, x_2, \ldots, x_n\), the expected value, \(E(X)\), also called the mean, is:$$\begin{aligned}E(X) &= \sum_{i=1}^{n} x_i P(X = x_i)\\ &= x_1 P(X = x_1) + x_2 P(X = x_2) + \cdots + x_n P(X = x_n)\\ \end{aligned}$$

You toss 2 fair coins, and \(X\) is the number of heads. The probability distribution is:

\(x\)	0	1	2
\(P(X = x)\)	\(\frac{1}{4}\)	\(\frac{1}{2}\)	\(\frac{1}{4}\)

Find the expected value of \(X\).

For any random variable \(X\) and constants \(a\) and \(b\), the expected value of a linear transformation of \(X\) is:$$ E(aX + b) = aE(X) + b $$This property is derived from two simpler rules:

• \(E(aX) = aE(X)\) (The expectation of a scaled variable is the scaled expectation).
• \(E(X+b) = E(X) + b\) (The expectation of a shifted variable is the shifted expectation).

The variance measures how spread out the values of a random variable are from its expected value. The standard deviation is the square root of the variance, giving a sense of typical deviation in the same units as \(X\).

The variance, denoted \(V(X)\), is:$$\begin{aligned}V(X) &= \sum_{i=1}^{n} (x_i - E(X))^2 P(X = x_i)\\ &= \left(x_1-E(X)\right)^2 P(X = x_1) + \left(x_2-E(X)\right)^2 P(X = x_2) + \cdots + \left(x_n-E(X)\right)^2 P(X = x_n)\\ \end{aligned}$$The standard deviation, denoted \(\sigma(X)\), is \(\sigma(X) = \sqrt{V(X)}\).

You toss 2 fair coins, and \(X\) is the number of heads. The probability table is:

\(x\)	0	1	2
\(P(X = x)\)	\(\frac{1}{4}\)	\(\frac{1}{2}\)	\(\frac{1}{4}\)

Given \(E(X) = 1\), find the variance.

A more convenient formula for computation is:$$V(X) = E(X^2) - [E(X)]^2$$

A random variable \(X\) follows a uniform distribution if each possible value has the same probability:$$P(X = x) = \frac{1}{\text{Number of possible values}}, \quad \text{for any possible value }x$$

Let \(X\) be the result of rolling a fair die: .

1. List the possible values of \(X\).
2. Create the probability table.
3. Draw the probability distribution graph.

For a random variable \(X\) that follows a uniform distribution on the set of integers \(\{1, 2, \ldots, n\}\):

• The expected value is \(E(X) = \frac{n+1}{2}\).
• The variance is \(V(X) = \frac{n^2-1}{12}\).

Let \(X\) be the random variable for the score on a roll of a fair six-sided die. Find the mean and variance of \(X\).

A Bernoulli distribution models an experiment with two outcomes: success (1) or failure (0), like flipping a coin where heads is 1 and tails is 0. The probability of success is \(p\).

A random variable \(X\) follows a Bernoulli distribution if:

• Possible values are 0 and 1.
• \(P(X = 1) = p\) and \(P(X = 0) = 1 - p\).

We write \(X \sim B(p)\).

A basketball player has an 80\(\pourcent\) chance of making a free throw. Let \(X = 1\) if the shot is made, and \(X = 0\) if it’s missed.

1. Is \(X\) a Bernoulli random variable?
2. Find the probability of success.

For a Bernoulli random variable \(X\) with a probability of success \(p\), the following hold:

• The expected value is \(E(X) = p\),
• The variance is \(V(X) = p(1 - p)\),
• The standard deviation is \(\sigma(X) = \sqrt{p(1 - p)}\).

The repetition of \(n\) identical and independent Bernoulli trials is called a Bernoulli scheme of size \(n\).

We toss an unfair coin twice where the probability of success (landing on HEADS) is \(0.4\).
Let \(X\) be the random variable that counts the number of HEADS. Find \(P(X=1)\).

In a binomial experiment (Bernoulli scheme) of \(n\) trials, the binomial coefficient \(\binom{n}{k}\), read as "\(n\) choose \(k\)", represents the number of different ways (or paths) to obtain exactly \(k\) successes.

Choosing 0 successes or choosing \(n\) successes can only be done in a single way. So$$ \binom{n}{0} = 1 \quad \quad \binom{n}{n} = 1 $$

Choosing \(k\) successes is the same as choosing \(n-k\) failures. Therefore:$$ \binom{n}{k} = \binom{n}{n-k} $$

For any integers \(1 \leq k \leq n-1\):$$ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} $$

This relationship allows us to construct binomial coefficients step-by-step using Pascal's Triangle.

To find \(\binom{n}{k}\), we sum the two coefficients directly above it in the previous row.

\(n \backslash k\)	0	1	2	3	4	5
0	1
1	1	1
2	1	2	1
3	1	3	3	1
4	1	4	6	4	1
5	1	5	10	10	5	1

Using Pascal's Triangle, find \(\binom{5}{2}\).

Let \(X\) be a binomial random variable with \(n\) independent trials and a probability of success \(p\). The probability distribution of \(X\) is:This is called the binomial distribution, and we write \(X \sim B(n, p)\).

A basketball player has an 80\(\pourcent\) chance of making a free throw and takes 5 shots. Let \(X\) be the number of shots made.

1. Is \(X\) a binomial random variable?
2. Find the probability of making 4 shots.

For \(X \sim B(n, p)\):

• \(E(X) = n p\) (expected value),
• \(V(X) = n p (1 - p)\) (variance),
• \(\sigma(X) = \sqrt{n p (1 - p)}\) (standard deviation).

A basketball player has an 80\(\pourcent\) chance of making a free throw and takes 5 shots. Find the mean and standard deviation of the number of successful shots.

To calculate probabilities of the form \(P(X \leq k)\), we use the Cumulative Distribution Function (CDF) on a scientific calculator.

• On TI: Use \texttt{binomcdf(n, p, k)}.
• On Casio: Use \texttt{BinomialCD(k, n, p)}.
• On NumWorks: Use the \texttt{Probability} application, select \texttt{Binomial}.

To handle other inequalities, use these logic rules:

• \(P(X < k) = P(X \leq k-1)\)
• \(P(X \geq k) = 1 - P(X \leq k-1)\)
• \(P(X > k) = 1 - P(X \leq k)\)
• \(P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)\)

Consider a random variable \(X\) that follows the binomial distribution with parameters \(n = 100\) and \(p = 0.78\). Calculate the following probabilities. Round your results to three decimal places. $$P(X < 75), P(X > 79), P(X \geq 74), P(73 < X \leq 81)$$

To find an interval \(I=[a ; b]\) such that \(P(X \in I) \geq 1-\alpha\):

1. Find the smallest integer \(a\) such that \(P(X \leq a) > \frac{\alpha}{2}\).
2. Find the smallest integer \(b\) such that \(P(X \leq b) \geq 1 - \frac{\alpha}{2}\).

Let \(X \sim \mathcal{B}(50, 0.4)\). Find the smallest interval \([a, b]\) such that \(P(a \leq X \leq b) \geq 0.95\).

Consider a sequence of independent Bernoulli trials with a probability of success \(p\). The geometric distribution models the number of trials \(X\) required to obtain the first success. We write \(X \sim \mathcal{G}(p)\).

Let \(X \sim \mathcal{G}(p)\). For any integer \(k \geq 1\):

• \(P(X = k) = (1-p)^{k-1}p\)
• \(P(X \leq k) = 1 - (1-p)^k\)
• \(P(X > k) = (1-p)^k\)
• \(E(X) = \frac{1}{p}\)

We roll a fair four-sided die (numbered 1 to 4) until we get a 2. Let \(D\) be the number of trials required.

1. Identify the distribution of \(D\).
2. Find the probability that the first 2 appears on the 5th trial.
3. Find the expected number of rolls to get a 2.

The bar chart of a geometric distribution shows an exponential decrease. The height of the first bar (at \(k=1\)) is equal to \(p\).

The probability distribution graph of \(X \sim \mathcal{G}(0.25)\) is:

For a random variable \(X\) following a geometric distribution, the probability of success in future trials does not depend on the number of past failures:$$ P_{X>s}(X > s+t) = P(X > t) \quad \text{for all } s, t \in \mathbb{N} $$

Following the previous die example (\(p=0.25\)), find the probability that it takes more than 10 trials to get a 2, given that after 7 trials, no 2 has appeared.