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Discrete Mathematics Unit 5: Discrete probability

Discrete probability counts favorable outcomes over total outcomes, using the counting tools from the previous unit.

Covers: Sample spaces, conditional probability, expected value

Study guide

Discrete probability works with finite or countable sample spaces. The sample space S is the set of all possible outcomes, and an event E is a subset. When outcomes are equally likely, P(E) = |E|/|S|, so counting techniques (permutations and combinations) do most of the work. Probabilities are between 0 and 1, P(Eᶜ) = 1 − P(E), and P(E ∪ F) = P(E) + P(F) − P(E ∩ F).

Conditional probability is P(A | B) = P(A ∩ B)/P(B): the probability of A once you know B happened. Events are independent if P(A ∩ B) = P(A)P(B). Bayes' theorem reverses conditioning: P(A | B) = P(B | A)P(A)/P(B), where P(B) often comes from the law of total probability, summing over cases. It explains results like why a positive test for a rare disease can still be likely to be a false positive.

A random variable assigns a number to each outcome. Its expected value is E(X) = Σ x·P(X = x), the long-run average; a fair die has E(X) = 3.5. Linearity of expectation says E(X + Y) = E(X) + E(Y) even when X and Y are dependent, which makes many problems easy by splitting a count into indicator variables. Variance measures spread: V(X) = E(X²) − E(X)².

Bernoulli trials (independent success/failure trials) lead to the binomial distribution: P(k successes in n trials) = C(n, k)pᵏ(1 − p)ⁿ⁻ᵏ.

For a nonnegative random variable X and any a > 0, P(X ≥ a) ≤ E(X)/a. It needs only the mean. Shifting X by a known lower bound first often gives a sharper result. For any random variable with mean μ and variance σ², P(|X − μ| ≥ c) ≤ σ²/c². This bounds the chance of being far from the mean using only the variance.

Var(aX + b) = a²·Var(X), and Var(X + Y) = Var(X) + Var(Y) when X and Y are independent. So the average of n independent copies of X has variance Var(X)/n, and the standard deviation is the square root of the variance. If independent trials each succeed with probability p, the number of trials until the first success has a geometric distribution with expected value 1/p.

Key ideas

Formulas and rules

P(E) = |E|/|S|
Bayes: P(A | B) = P(B | A)P(A)/P(B)
E(X + Y) = E(X) + E(Y)

Common mistakes

Words to know

Probability with equally likely outcomes
P(E) = |E|/|S|
Bayes' theorem
P(A | B) = P(B | A)P(A)/P(B)
Linearity of expectation
E(X + Y) = E(X) + E(Y), even if X and Y are dependent.
"At least one" probability
1 − P(none)
Expected value
E(X) = Σ x·P(x); a fair die averages 3.5.
Sample space
The set of all possible outcomes.
Event
A subset of the sample space.
Conditional probability
P(A | B) = P(A ∩ B)/P(B)
Independent events
P(A ∩ B) = P(A)P(B)
Law of total probability
P(B) = Σ P(B | Aᵢ)P(Aᵢ)
Variance
V(X) = E(X²) − E(X)²
Bernoulli trial
An independent trial with two outcomes.
Indicator variable
1 if an event happens, 0 if not; its expected value is the probability.

Sample practice questions

27 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

P(sum of 7) with two dice.

  1. 7/36
  2. 1/6
  3. 1/12
  4. 1/36
Show the answer and steps

Answer: B) 1/6

  1. Sums of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
  2. 6/36 = 1/6.
Question 2 (level 1)

Expected number of heads in 47 fair coin flips?

  1. 47
  2. 1/2
  3. 47/2
  4. 47/4
Show the answer and steps

Answer: C) 47/2

  1. Linearity of expectation: each flip contributes 1/2.
  2. 47 × 1/2 = 47/2.
Question 3 (level 3)

3 red and 2 blue marbles; draw 2 without replacement. P(both red)?

  1. 3/10
  2. 1/2
  3. 3/5
  4. 9/25
Show the answer and steps

Answer: A) 3/10

  1. P(first red) = 3/5, P(second red) = 2/4.
  2. 3/5 × 2/4 = 3/10.