Independent Events and Unions of Events: Study Guide (Statistics 2.7)

What does “independent” mean?

Two events are independent if knowing whether one of them happened does not change the probability of the other. Learning about A gives you no new information about B.

Think of two separate spins of a fair spinner. The first spin landing on 3 tells you nothing about the second spin. The events “first spin is 3” and “second spin is 3” are independent.

Now think of one person chosen at random from a school. “The person is over 180 cm tall” and “the person plays basketball for the school team” are probably not independent. If you learn that the person plays basketball, the probability that they are tall goes up.

Independence is about probabilities, not about whether the events feel connected. You decide it with numbers, or you state it as an assumption when the situation justifies it.

Three ways to check for independence

From Topic 2.6, P(A | B) is the probability of A given that B has happened. Events A and B are independent if and only if any one of these is true:

Condition	In words
P(A | B) = P(A)	Knowing B happened does not change the probability of A.
P(B | A) = P(B)	Knowing A happened does not change the probability of B.
P(A ∩ B) = P(A) · P(B)	The probability of both equals the product of the separate probabilities.

The three conditions are equivalent (as long as P(A) and P(B) are not 0). If one holds, they all hold. If one fails, they all fail. So you need to check only one. Pick the one that uses numbers you already have.

Where the product rule comes from. The general multiplication rule from Topic 2.6 says P(A ∩ B) = P(A) · P(B | A). If A and B are independent, P(B | A) is just P(B). Replace it and you get

P(A ∩ B) = P(A) · P(B) (independent events only).

This is the multiplication rule for independent events. It extends to more than two events: if A, B and C are independent, P(A ∩ B ∩ C) = P(A) · P(B) · P(C).

How to write a check. Show both numbers and compare them. For example: “P(F | C) = 0.40 and P(F) = 0.46. Since 0.40 ≠ 0.46, the events are not independent.” A bare “not equal, so dependent” earns little.

The union of two events: A or B

The union A ∪ B is the event that A happens, or B happens, or both happen. In statistics, “or” always includes “both” unless you are told otherwise.

You might try P(A) + P(B). The problem is the overlap. Outcomes in both A and B are counted once inside P(A) and again inside P(B). To count them only once, subtract the overlap one time. This gives the general addition rule:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

This rule works for any two events. Two special cases:

• If A and B are mutually exclusive (Topic 2.5), P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
• If A and B are independent, you can find the overlap with the product rule: P(A ∪ B) = P(A) + P(B) − P(A) · P(B).

A useful check: P(A ∪ B) can never be bigger than 1, and can never be smaller than the larger of P(A) and P(B). Also, P(A ∪ B) = 1 − P(neither A nor B), because “neither” is the complement of “A or B”.

Seeing a union on a Venn diagram

The data below come from a fictional ferry company, Northwind Ferries. On one crossing, 800 passengers were recorded. Event C: the passenger travelled with a car. Event F: the passenger bought food on board.

The same counts as a two-way table:

	Bought food (F)	No food	Total
Car (C)	128	192	320
No car	240	240	480
Total	368	432	800

Worked example 1: checking independence and finding a union from a table

Question. One passenger is chosen at random from the 800. (a) Find P(C), P(F) and P(C ∩ F). (b) Are C and F independent? Justify your answer. (c) Find P(C ∪ F) and interpret it.

(a)

1. P(C) = 320 ÷ 800 = 0.40.
2. P(F) = 368 ÷ 800 = 0.46.
3. P(C ∩ F) = 128 ÷ 800 = 0.16.

(b) Use the conditional-probability check. Among car passengers, P(F | C) = 128 ÷ 320 = 0.40. For all passengers, P(F) = 0.46. Since 0.40 ≠ 0.46, knowing that a passenger has a car changes the probability that they bought food. C and F are not independent.

The product check gives the same verdict: P(C) · P(F) = 0.40 × 0.46 = 0.184, but P(C ∩ F) = 0.16. Since 0.184 ≠ 0.16, the events are not independent.

(c) P(C ∪ F) = P(C) + P(F) − P(C ∩ F) = 0.40 + 0.46 − 0.16 = 0.70.

Interpretation. If one of these 800 passengers is chosen at random, the probability that the passenger travelled with a car, bought food on board, or both, is 0.70.

Check. The only passengers outside C ∪ F are the 240 with no car and no food. 1 − 240 ÷ 800 = 1 − 0.30 = 0.70. It matches. Without the subtraction you would get 0.86, which counts the 128 “both” passengers twice.

Note. Car passengers were less likely to buy food (0.40 compared with 0.50 for passengers without a car). The data do not tell us why. Perhaps drivers ate in their cars.

Worked example 2: independent events and “at least one”

Question. A fictional bakery, Crumb & Co., has an oven and a mixer. On any working day, the oven breaks down with probability 0.04 and the mixer breaks down with probability 0.07. Assume breakdowns of the two machines are independent.

(a) Find the probability that both machines break down on the same day. (b) Find the probability that at least one machine breaks down on a given day. (c) Find the probability that there is at least one day with a breakdown in a 5-day working week. Assume different days are independent.

(a) Let O = oven breaks down, M = mixer breaks down. Because they are independent,

P(O ∩ M) = P(O) · P(M) = 0.04 × 0.07 = 0.0028.

(b) “At least one” is the union O ∪ M.

P(O ∪ M) = P(O) + P(M) − P(O ∩ M) = 0.04 + 0.07 − 0.0028 = 0.1072.

Check with the complement. “At least one breaks down” is the opposite of “neither breaks down”. P(neither) = (1 − 0.04)(1 − 0.07) = 0.96 × 0.93 = 0.8928. Then 1 − 0.8928 = 0.1072. It matches. (Multiplying 0.96 and 0.93 is allowed because if O and M are independent, so are their complements.)

(c) On one day, P(no breakdown) = 0.8928. For 5 independent days,

P(no breakdown on any of the 5 days) = 0.8928⁵ ≈ 0.5672.

P(at least one day with a breakdown) = 1 − 0.5672 ≈ 0.4328.

Interpretation. In about 43% of 5-day weeks, the bakery can expect at least one day on which a machine breaks down.

Why not 5 × 0.1072? That gives 0.536, which counts weeks with two or more breakdown days more than once. With more days the same mistake gives a “probability” above 1. The complement route avoids this.

Is independence reasonable? Only if the machines do not share a cause of failure. If both run on the same electrical circuit, a power cut could stop both, and the multiplication rule would underestimate P(O ∩ M). State the assumption and think about whether it fits.

Worked example 3: independent is not the same as mutually exclusive

Question. A fair spinner has 10 equal sectors numbered 1 to 10. It is spun once. Let A = “the number is even”, B = “the number is 4 or less” and C = “the number is 9 or 10”. (a) Are A and B independent? (b) Are B and C independent? Are they mutually exclusive? (c) Find P(A ∪ B) and P(B ∪ C).

(a) A = {2, 4, 6, 8, 10}, so P(A) = 0.5. B = {1, 2, 3, 4}, so P(B) = 0.4. A ∩ B = {2, 4}, so P(A ∩ B) = 0.2.

P(A) · P(B) = 0.5 × 0.4 = 0.2 = P(A ∩ B). So A and B are independent. In words: P(A | B) = 2 ÷ 4 = 0.5 = P(A). Knowing the number is 4 or less leaves the chance of an even number at one half.

(b) C = {9, 10}, so P(C) = 0.2. B and C share no outcomes, so P(B ∩ C) = 0: they are mutually exclusive.

P(B) · P(C) = 0.4 × 0.2 = 0.08, but P(B ∩ C) = 0. Since 0 ≠ 0.08, B and C are not independent. Knowing B happened tells you C is impossible: P(C | B) = 0, not 0.2.

(c) P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7. Check by listing: A ∪ B = {1, 2, 3, 4, 6, 8, 10}, which is 7 of 10 sectors.

P(B ∪ C) = 0.4 + 0.2 − 0 = 0.6. Check: {1, 2, 3, 4, 9, 10}, 6 of 10 sectors.

The lesson. Mutually exclusive events with non-zero probabilities are always dependent: if one happens, the other cannot. Independent events with non-zero probabilities always overlap.

Common misconceptions

• “Independent means mutually exclusive.” These are different ideas. Mutually exclusive means P(A ∩ B) = 0. Independent means P(A ∩ B) = P(A) · P(B). Both can be true only if one of the events has probability 0.
• Adding for “or” when the events overlap. P(A) + P(B) works only for mutually exclusive events. Otherwise subtract P(A ∩ B). A “probability” above 1 is a warning sign.
• Multiplying without checking independence. P(A) · P(B) gives P(A ∩ B) only for independent events. Otherwise use P(A) · P(B | A).
• “The events have nothing to do with each other, so they must be independent.” Independence must be checked with data or justified by how the outcomes are produced. State it as an assumption when you use it.
• “After four reds in a row, a different colour is due.” For independent spins, past results do not change the next spin. The probability stays the same each time.
• “P(at least one in n tries) = n × p.” This overcounts. Use 1 − P(none).
• Comparing the wrong pair. To check independence with conditional probabilities, compare P(A | B) with P(A), not with P(B), and not P(A | B) with P(B | A).
• A verdict with no numbers. Write both values and the comparison: “0.40 ≠ 0.46, so not independent”.

Where this leads

Next, in Topic 2.8, you will use these rules to build probability distributions: tables that give the probability of each possible value of a random variable. Independence and the multiplication rule are how many of those probabilities are found, and they return in the binomial distribution in Topic 2.10. Read Introduction to Random Variables and Probability Distributions when you are ready.

First, try the practice questions, then use the revision notes and the checklist to consolidate. If conditional probability still feels shaky, go back to Conditional Probability.