Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From lesson 3: P(A) = 0.7, P(B) = 0.4 and P(A ∩ B) = 0.2. Find P(A ∪ B).
Answer
0.9. 0.7 + 0.4 − 0.2: the 0.2 was counted twice. - From Year 8: a fair coin is flipped twice. What is the probability of two heads?
Answer
1/4. 1/2 × 1/2, or one of four equally likely outcomes. - From Year 9: expand p(1 − q).
Answer
p − pq.
Words used in this lesson
- independent events
- events where one happening does not change the chance of the other. For example: a head on one coin and a head on a second coin.
- mutually exclusive
- events that cannot both happen. For example: a die showing 1 and the same die showing 6.
- dependent events
- events that are not independent. For example: drawing two cards from one pack without putting any back.
- product
- the result of multiplying. For example: 0.8 × 0.4 = 0.32.
- overlap
- the outcomes that are in both events. For example: A ∩ B.
- test for independence
- comparing P(A ∩ B) with P(A) × P(B). For example: 0.08 against 0.2 × 0.3.
Sleep and screens
A school nurse in Auckland records two facts about each student in a survey. In the whole survey, 30% of students slept fewer than 8 hours, and 40% used a screen after 22:00 on school nights.
Predict first: if the two facts had nothing to do with each other, what share of students would be in both groups? What would it look like if late screens made short sleep more likely? Test your ideas on the figure: move the overlap and see when it matches what chance alone would give.
P(A) = 80%, P(B) = 50%: if independent the overlap would be 40%. The overlap is 30%, less than the product, so A and B are not independent.
Show me the steps
- Start at P(A) = 80%, P(B) = 50% and P(A ∩ B) = 30%. Chance alone would give 80% of 50% = 40%, but the overlap is 30%, which is less, so A and B are not independent.
- Set P(A ∩ B) to 40%: the overlap equals P(A) × P(B), and the B column cuts the A strip at the same place as it cuts the whole square. A and B are independent.
- Set P(A) = 50%, P(B) = 30% and P(A ∩ B) = 0%: the events are mutually exclusive, the product is 15%, and 0% is not 15%. Mutually exclusive events with positive probabilities are never independent.
- Now set P(A ∩ B) to 15%: 50% of 30% is 15%, so they are independent even though the overlap is a small share of the whole.
Independent events. A and B are independent when P(A ∩ B) = P(A) × P(B). Then knowing that A happened does not change the chance of B: the share of the A strip that is B equals the share of the whole that is B.
Not the same as mutually exclusive. Mutually exclusive means P(A ∩ B) = 0. Independent means P(A ∩ B) = P(A) × P(B). If both events can happen, mutually exclusive events are never independent, because one happening rules the other out.
To test, find P(A ∩ B) first, from the data or from the addition rule, and compare it with the product. Equal means independent; anything else means dependent.
“Mutually exclusive” and “independent” are not the same. Roll one fair die with A = “a 1” and B = “a 6”. They are mutually exclusive, P(A ∩ B) = 0. But P(A) × P(B) = 1/6 × 1/6 = 1/36, which is not 0, so they are dependent: if A happens, B cannot.
Worked examples
1. P(A) = 0.4, P(B) = 0.15 and P(A ∪ B) = 0.49. Are A and B independent?
- P(A ∩ B) = 0.4 + 0.15 − 0.49 = 0.06.
- P(A) × P(B) = 0.4 × 0.15 = 0.06.
- They are equal, so A and B are independent.
2. P(A) = 0.3 and P(B) = 0.2, and A and B are mutually exclusive. Are they independent?
- Mutually exclusive means P(A ∩ B) = 0.
- P(A) × P(B) = 0.3 × 0.2 = 0.06, which is not 0.
- So they are not independent.
3. A and B are independent. Prove that A and B′ are independent.
- P(A ∩ B′) = P(A) − P(A ∩ B), because A is split into A ∩ B and A ∩ B′.
- By independence P(A ∩ B) = P(A) × P(B), so P(A ∩ B′) = P(A) − P(A) × P(B) = P(A) × (1 − P(B)).
- By the complement rule 1 − P(B) = P(B′), so P(A ∩ B′) = P(A) × P(B′).
Your go. The first step is given. A and B are independent. Prove that A′ and B are independent.
- P(A′ ∩ B) = P(B) − P(A ∩ B), because B is split into A ∩ B and A′ ∩ B.
- Now substitute the independence test, factorize, and use the complement rule.
Check the answer
P(A′ ∩ B) = P(B) − P(A) × P(B) = P(B) × (1 − P(A)) = P(A′) × P(B), so A′ and B are independent.
Check before you go on
1. P(A) = 0.5, P(B) = 0.2 and P(A ∩ B) = 0.1. Which statement is correct?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. A and B are independent with P(A) = 0.45 and P(B) = 0.6. Find P(A ∩ B).
3. A and B are independent, P(A) = 0.4 and P(A ∪ B) = 0.82. Find P(B).
Level 2 · Building
4. P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.85. How much smaller is P(A ∩ B) than P(A) × P(B)?
5. A and B are independent, P(A) = 0.3 and P(B) = 0.8. Find P(A′ ∩ B).
Level 3 · Stretch
6. A and B are independent. P(A) = 0.3 and P(neither A nor B) = 0.28. Find P(B).
7. A and B are mutually exclusive and also independent, and P(A) = 0.5. Find P(B).
Something to think about
A and B are independent. Is it possible for A and B to be the same event with a probability strictly between 0 and 1? Try P(A ∩ A) = P(A) × P(A). Then say which events are independent of themselves.
One way to think about it
If B is the same event as A, then A ∩ B = A, so independence needs P(A) = P(A) × P(A), that is P(A) × (1 − P(A)) = 0. So P(A) is 0 or 1. An event is independent of itself only when it is impossible or certain: knowing the event has happened tells you something unless there was nothing left to learn.
Exit ticket
8. One fair die is rolled. Which pair of events is independent?