Feb 18, 2026

IT 223 -- Feb 18, 2026

Review Exercises

What is a discrete random variable?
Answer: It is a random variable with only a finite number of outcomes. Contrast this with a normal random variable, which can output any value in the range (-∞, +∞)
What is the expected value of a discrete random variable?
Answer: If x₁, x₂, ... , x_n are the values of the random variable x and p₁, p₂, ... , p_n are the corresponding probilities, the expected value of X is
E(X) = x₁ p₁ + x₂ p₂ + ... + x_n p_n
Compute E(x) using R like this:
```
> values <- c(x1, x2, x3, x4, x5)
> probs <- c(p1, p2, p3, p4, p5)
> expectedValue <- sum(values * probs)
```
What is the variance of a discrete random variable?
Answer: if P(X=x_i) = p_i for each i,
Var(X) = (x_i - E(X))² * p₁ + ... + (x_n - E(X))² * p_n
Compute the variance like this:
```
variance <- sum((values - expected_value)^2 * probs)
```
How is the variance related to the standard deviation of a dataset or of a random variable?
Answer: The variance is the square of the standard deviation.
What is the expected value of a Bernoulli random variable?
Answer: for a Bernoulli random variable, P(X = 0) = 1 - p and P(X = 1) = p. Then
E(X) = 0 * (1 - p) + 1 * p = p, so that E(X) = p.
What is the variance of a Bernoulli random variable?
Answer: Var(X) = (x₁ - E(X))² p₁ + ... + (x_n - E(X))² p_n
   = (0 - p)² (1 - p) + (1 - p)² p
   = p² (1 - p) + (1 - 2p + p²) p
   = p² - p³ + p - 2p² + p³
   = p - p² = p(1 - p), so
Var(X) = p (1 - p)
If S is the sum of n independent outcomes x₁, ... , x_n from a random variable X, what are E(S) and Var(S) in terms of E(X) and Var(X)?
Answer: E(S) = E(X₁ + ... + X₁) = E(X₁) + ... + E(X_n) = nE(X), so E(S) = nE(X)
Undergrads at a university have these probabilities for number of times changing majors:
Times: 0 1 2 3 4 5
Probability: 0.28 0.37 0.23 0.09 0.02 0.01
1. expected value of the number of major changes for a randomly selected student at this university,
  Answer: We can use R to compute the expected value like this:
```
> times <- 0:5
> times
[1] 0 1 2 3 4 5
> prob <- scan( )
1: 0.28 0.37 0.23 0.09 0.02 0.01
7: 
Read 6 items
> prob
[1] 0.28 0.37 0.23 0.09 0.02 0.01
> # The next line computes the expected value:
> sum(times * prob)
[1] 1.23
```
2. variance of the number of major changes for a student,
```
> # Answer: The variance is computed like this:
> sum((times - 1.23)^2 * prob)
[1] 1.1571
```
3. standard deviation of the number of major changes at a university,
```
> # Answer: the SD is the square root of the variance:
> sqrt(1.1571)
[1] 1.075686
```
If A and B are two events, what must we assume for
P(A and B) = P(A) * P(B)
to be true?
Answer: A and B must be independent.
If A and B are two events, what must be true for
P(A or B) = P(A) + P(B) to be true?
Answer: A and B must be mutually exclusive.
When rolling a pair of dice, snakeeyes are double ones. What is the probability of rolling at least one pair of snakeeyes in twenty rolls of two dice?
Answer: 1 - (1 - 1/36)²⁰ = 0.4307 - 43%. We can use R to perform the calculation like this:
```
> 1 - (1 - 1/36)^20
[1] 0.4307397
```
If the probability of getting a cold in a given week is 3%, what is the probability of getting at least one cold in a year (52 weeks)?
Answer: 1 - (1 - p)ⁿ = 1 - (1 - 0.03)⁵² = 0.7948 = 79.5%.
We can use R to perform the calculation like this:
```
> 1 - (1 - 0.03)^52
[1] 0.7948223
```

Properties of Random Variables

c is a constant. A constant can be thought of as a random variable with this probability distribution:

Outcome	Probability
c	1

E(c) = c
E(cx) = c E(x)
E(x + y) = E(x) + E(y)
E(x₁ + ... + x_n) = E(x₁) + ... + E(x_n)
Var(x) = E(x²) - E(x)²
Definition: x and y are independent if E(xy) = E(x)E(y).
If x and y are independent, Var(x + y) = Var(x) + Var(y)
Var(cx) = c² Var(x)

Here are the derivations.

Practice Problem

Compute the variance and SD of a Bernoulli random variable with the formula Var(x) = E(x²) - E(x)².
Var(x) = E(x²) - E(x)² = (0² (1 - p) + ... + 1² p) - p²
= p - p² = p(1 - p)
This is the same result that we obtained earlier.

Sums of Random Variables

Define S = x₁ + ... + x_n, where each x_i has the same expected value and variance, and all of the x_i are independent.
The expected value of the sum of n outcomes of the random variable x:
E(S) = E(x₁ + ... + x_n) = E(x₁) + ... + E(x_n) = nE(x)

E(S) = nE(x)
The variance of the sum of n independent outcomes of the random variable x:
Var(S) = Var(x₁ + ... + x_n) = Var(x₁) + ... + Var(x_n) = nVar(x)

Var(S) = nVar(x)
The standard deviation of the sum of n independent outcomes of a random variable:
σ_S = σ_{x₁ + ... + x_n} = √[Var(x₁ + ... + x_n)]
= √[Var(x₁) + ... + Var(x_n)] = √nVar(x) = σ_x√n

σ_S = σ_x√n

Practice Problems

Compute the expected value and standard deviation of the total amount of rain in a year (365 days) if the number of inches of rain in one day is this random variable:
Amount: 0 1 2 3
Probability: 0.3 0.4 0.2 0.1
Compute the expected value and standard deviation of independent Bernoulli random variable outcomes for the sum of n trials.

Using R to Simulate Bernoulli Outcomes

Use R to simulate the random outcomes from flipping a fair coin 10 times. Answer: the probability experiment is flipping a coin once; this experiment is repeated 10 times. A binomial random variable is the sum of a specified number of independent outcomes from a Bernoulli random variable. We will see why this type of random variable is called a binomial random variable later.
```
// Here are the meanings of the R arguments:
// n = number of times the experiment is repeated
// size = number of Bernoulli outcomes in one experiment
// prob = the probability of success for Bernoulli rv.
rbinom(n=10, size=1, prob=0.5)
 [1] 1 0 0 1 1 1 1 0 0 1
```
Use R to simulate 20 repetitions of flipping a coin 2 times.
Answer: in this case, n = 20, size = 2, and prob = 0.5.
```
> rbinom(n=20, size=2, prob=0.5)
# or
> rbinom(20, 2, 0.5)
 [1] 1 2 1 2 0 1 1 1 0 2 0 2 2 2 0 1 1 1 1 1
```
Use R to simulate flipping a fair coin one million times. Repeat the experiment twice. Answer:
```
> > rbinom(2, 1000000, 0.5) 
[1] 500664 499815
```
Elena Delle Donne is the best free throw shooter in WNBA history with a career average of 93.4%. Simulate 10 seasons of shooting 90 free throws per season with a success rate of 93.4%. Answer:
```
> rbinom(10, 90, 0.934)
[1] 79 82 84 87 86 81 85 86 82 87
```

Averages of Random Variables

We will discuss this section later.
The expected value of the average of n outcomes of a random variable:
E(x) = E[(x₁ + ... + x_n) / n] = E(x₁ / n + ... + x_n / n)
= E(x₁ / n) + ... + E(x_n / n) = E(x₁) / n + ... + E(x_n) / n
= nE(x) / n = E(x)

E(x) = E(x)
The variance of the sum of n independent outcomes of a random variable:
Var(x) = Var[(x₁ + ... + x_n) / n] = Var[x₁ / n + ... + x_n / n]
= Var(x₁ / n) + ... + Var(x_n / n) = Var(x₁) / n² + ... + Var(x_n) / n²
= nVar(x) / n² = Var(x) / n

Var(x) = Var(x) / n
The standard deviation of the sum of n independent outcomes of a random variable:
σ_x = √[Var(x)] = √[Var(x) / n] = σ_x / √n

σ_x = σ_x / √n
Recall that E(x) = E(x). How does σ_x compare to σ_x?

The Law of Averages

The official name of the Law of Averages is the Law of Large Numbers (LLN). It was first proved by Jakob Bernoulli from Basil, Switzerland in 1713.
LLN states that if x₁, ... , x_n are independent outcomes of the same random variable x and n is large enough, then the average x is close to the expected value E(x) = E(x) with high probability.
LLN works by swamping, not by compensation. What does this mean?
Ans: It means that if the average of outcomes of a random variable is larger than the expected value of the average, there is no reason to expect that if you continue to collect outcomes from the random variable that those new outcomes will be smaller than the expected value of the average to compensate. Instead it means that by swamping the average with many more outcomes, the new outcomes will be close to the average with high probability, so the overall average will also tend to be closer to the expected value. Note that this is true with high probability; there will always be a small probability that the overall average is not close to the expected value of the average.
Experiment: Use R to simulate 100, 1000, and 10,000 tosses of a fair coin. Are the averages obtained consistant with LLN?
Answer: Here are the results:

n S x

       1,000       469 0.4690

     10,000    4,925 0.4925

   100,000   49,781 0.4978

1,000,000 500,107 0.5001

n	S	x
1,000	469	0.4690
10,000	4,925	0.4925
100,000	49,781	0.4978
1,000,000	500,107	0.5001

Factorials and Counting Combinations

Recall that n! (read n factorial, n bang) is defined as
n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1
n! is the number of ways of arranging n distict objects.
3! = 6. For example, the ways of arranging A, B, and C is
(ABC) (ACB) (BAC) (BCA) (CAB) (CBA)
2! = 2. The ways of arranging A and B is
(AB) (BA)
1! = 1. The ways of arranging the single object A is
(A)
What is 0!?
Answer: There is exactly one way of arranging zero objects: ( ). Thus 0! = 1.
One concept that is important for probability is the number of ways of choosing k items from a total of n items (n choose k).
1. The number of ways to choose 1 item from 3 items:
  (1) (2) (3)
  Thus there are 3 ways to choose 1 item from 3 items. The R computation:
```
> choose(3, 1)
[1] 3
```
2. The number of ways to choose 2 items from 3 items:
  (12) (13) (23)
  Thus there are 3 ways to choose 2 items from 3 items. The R computation:
```
> choose(3, 2)
[1] 3
```
3. The number of ways to choose 2 items from 4 items:
  (12) (13) (14) (23) (24) (34)
  Thus there are 3 ways to choose 2 items from 4 items. The R computation:
```
> choose(4, 2)
[1] 6
```