Discrete Random Variables

A random variable, denoted \(X\), is a function that assigns a numerical value to each outcome in a random experiment.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 \(X(\textcolor{colordef}{H},\textcolor{colorprop}{T})\).

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 possibles 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 \(X\).

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 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.

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 \(\sigma(X) = \sqrt{p(1 - p)}\).

Suppose a basketball player takes \(n\) free throws, and we count the number of shots made. The probability of making a free throw is the same for each attempt, and each shot is independent of every other shot. This is an example of a binomial experiment.

In a binomial experiment:

• There are a fixed number of independent trials,
• Each trial has only two possible outcomes: success (if the event occurs) or failure (if it does not),
• The probability of success is constant for each trial.

Let \(X\) be the number of successes in a binomial experiment with \(n\) trials, each with a probability of success \(p\). \(X\) is called a binomial random variable.

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.