Probability and the binomial distribution: IB Maths AI Notes (4.3)

This note covers the rules of probability: complements, combined events, independence, conditional probability, and the use of Venn diagrams, tree diagrams and tables. It then introduces the binomial distribution, which models the number of successes in a fixed number of independent trials, and shows how to use your GDC to find binomial probabilities.

Key ideas

Basics. The sample space is the set of all possible outcomes. For an event \( A \), the probability is \( P(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of equally likely outcomes}} \), and \( 0 \le P(A) \le 1 \). The complement \( A' \) is "not \( A \)", with \( P(A') = 1 - P(A) \). The expected number of times an event happens in \( n \) trials is \( n \times P(A) \).

Combined events.

\( A \cap B \) means "\( A \) and \( B \)"; \( A \cup B \) means "\( A \) or \( B \) (or both)".
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). The subtraction avoids counting the overlap twice.
Events are mutually exclusive if they cannot happen together: \( P(A \cap B) = 0 \), so \( P(A \cup B) = P(A) + P(B) \).
Events are independent if one happening does not change the probability of the other: \( P(A \cap B) = P(A)\,P(B) \). To test independence, check whether this equation holds.
Mutually exclusive events with non-zero probabilities are never independent, because if one happens the other cannot.

Conditional probability. The probability of \( A \) given that \( B \) has happened is \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}. \] The "given" event is the new, reduced sample space. If \( A \) and \( B \) are independent, \( P(A \mid B) = P(A) \).

Diagrams. A Venn diagram shows how events overlap; fill in the overlap first. A tree diagram shows events in stages: multiply along the branches to find the probability of a path, and add the paths that make up the event you want. If items are chosen without replacement, the probabilities on the second stage change. A two-way table is useful for conditional probabilities: restrict yourself to the row or column of the given event.

The binomial distribution. Suppose an experiment is repeated \( n \) times with these conditions:

there is a fixed number \( n \) of trials;
each trial has just two outcomes, "success" and "failure";
the probability of success \( p \) is the same each time;
the trials are independent.

Then the number of successes \( X \) follows a binomial distribution, written \( X \sim \text{B}(n, p) \). The probability of exactly \( k \) successes is \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, \dots, n. \] The mean (expected value) is \( E(X) = np \) and the variance is \( \text{Var}(X) = np(1-p) \).

On your GDC, use the binomial probability density function for \( P(X = k) \) and the binomial cumulative distribution function for \( P(X \le k) \). For other probabilities, rewrite in terms of these:

\( P(X \ge k) = 1 - P(X \le k-1) \)
\( P(a \le X \le b) = P(X \le b) - P(X \le a-1) \)

Definitions and formulae

Term or formula	Meaning / when to use
\( P(A') = 1 - P(A) \)	Probability of "not \( A \)". Useful for "at least one".
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)	Probability of \( A \) or \( B \) (or both).
\( P(A \cap B) = P(A)P(B) \)	Holds exactly when \( A \) and \( B \) are independent.
\( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} \)	Probability of \( A \) given \( B \).
\( X \sim \text{B}(n, p) \)	Number of successes in \( n \) independent trials with success probability \( p \).
\( P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \)	Binomial probability. Use the GDC.
\( E(X) = np \), \( \text{Var}(X) = np(1-p) \)	Mean and variance of a binomial variable.

Worked examples

Example 1 In a group of 80 students, 45 study French (\( F \)), 30 study Spanish (\( S \)) and 12 study both. A student is chosen at random. Find \( P(F \cup S) \) and \( P(S \mid F) \), and decide whether \( F \) and \( S \) are independent.

\( P(F) = \dfrac{45}{80} \), \( P(S) = \dfrac{30}{80} \), \( P(F \cap S) = \dfrac{12}{80} \).

\( P(F \cup S) = \dfrac{45}{80} + \dfrac{30}{80} - \dfrac{12}{80} = \dfrac{63}{80} = 0.7875 \).

\( P(S \mid F) = \dfrac{P(F \cap S)}{P(F)} = \dfrac{12/80}{45/80} = \dfrac{12}{45} = \dfrac{4}{15} \approx 0.267 \).

Independence test: \( P(F)P(S) = \dfrac{45}{80} \times \dfrac{30}{80} \approx 0.211 \), but \( P(F \cap S) = \dfrac{12}{80} = 0.15 \). They are not equal, so \( F \) and \( S \) are not independent.

Answer: \( P(F \cup S) = \dfrac{63}{80} = 0.7875 \); \( P(S \mid F) = \dfrac{4}{15} \approx 0.267 \); not independent.

Example 2 A box has 5 red and 3 blue counters. Two are drawn at random without replacement. Find the probability that (a) both are the same colour, (b) at least one is blue, (c) the second is red given that the first is blue.

First draw: \( P(R) = \tfrac58 \), \( P(B) = \tfrac38 \). Second draw (one counter fewer): after \( R \): \( P(R) = \tfrac47 \), \( P(B) = \tfrac37 \). After \( B \): \( P(R) = \tfrac57 \), \( P(B) = \tfrac27 \).

(a) \( P(RR) = \tfrac58 \times \tfrac47 = \tfrac{20}{56} \), \( P(BB) = \tfrac38 \times \tfrac27 = \tfrac{6}{56} \). The paths are mutually exclusive, so add: \( \tfrac{26}{56} = \tfrac{13}{28} \approx 0.464 \).

(b) "At least one blue" is the complement of "both red": \( 1 - \tfrac{20}{56} = \tfrac{36}{56} = \tfrac{9}{14} \approx 0.643 \).

Answer: (a) \( \tfrac{13}{28} \approx 0.464 \); (b) \( \tfrac{9}{14} \approx 0.643 \); (c) \( \tfrac57 \approx 0.714 \).

Example 3 A machine makes components, and 8% of them are faulty, independently. A quality inspector selects 12 components. Let \( X \) be the number of faulty ones. Find (a) \( P(X = 2) \), (b) \( P(X \le 1) \), (c) \( P(X \ge 1) \), (d) the mean and variance of \( X \).

There are a fixed number of independent trials, two outcomes (faulty or not), and constant \( p = 0.08 \). So \( X \sim \text{B}(12, 0.08) \).

(a) Use the binomial probability density function on your GDC with \( n = 12 \), \( p = 0.08 \), \( k = 2 \): \( P(X = 2) \approx 0.183 \).

(b) Use the binomial cumulative distribution function with upper value 1: \( P(X \le 1) \approx 0.751 \).

(d) \( E(X) = np = 12 \times 0.08 = 0.96 \). \( \text{Var}(X) = np(1-p) = 12 \times 0.08 \times 0.92 = 0.8832 \approx 0.883 \).

Answer: (a) 0.183; (b) 0.751; (c) 0.632; (d) mean 0.96, variance 0.883.

Common errors

Common error: adding \( P(A) + P(B) \) for "A or B" without subtracting \( P(A \cap B) \) when the events can overlap.
Common error: confusing "mutually exclusive" with "independent". They are different ideas.
Common error: in a without-replacement tree, using the same probabilities on the second stage as on the first.
Common error: writing \( P(X \ge 4) = 1 - P(X \le 4) \). It should be \( 1 - P(X \le 3) \).
Common error: using a binomial model when trials are not independent or \( p \) changes (for example, drawing from a small group without replacement).

Exam tip: Start a binomial question by stating the distribution, \( X \sim \text{B}(n, p) \), and writing the probability you need in the form \( P(X = k) \) or \( P(X \le k) \). Then say which GDC function you use.

Check yourself

\( P(A) = 0.4 \), \( P(B) = 0.5 \) and \( P(A \cap B) = 0.2 \). Are \( A \) and \( B \) independent? Find \( P(A \cup B) \).
Answer
\( P(A)P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B) \), so they are independent. \( P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7 \).
A fair die is rolled 6 times. Find the probability of exactly two sixes.
Answer
\( X \sim \text{B}(6, \tfrac16) \). \( P(X = 2) = \binom62 \left(\tfrac16\right)^2\left(\tfrac56\right)^4 \approx 0.201 \).
A fair coin is flipped 10 times. Find the probability of at least 7 heads.
Answer
\( X \sim \text{B}(10, 0.5) \). \( P(X \ge 7) = 1 - P(X \le 6) = 1 - 0.828125 \approx 0.172 \).
\( X \sim \text{B}(20, 0.3) \). Find the mean, variance and \( P(X = 5) \).
Answer
Mean \( = 20 \times 0.3 = 6 \). Variance \( = 20 \times 0.3 \times 0.7 = 4.2 \). \( P(X = 5) \approx 0.179 \).