High School Maths · Year 2 · Unit 13 · Lesson 1 of 5

The AND and OR Rules: Are These Events Really Independent?

About 15 minutes · independent events and P(A and B) = P(A) × P(B) · mutually exclusive events · the addition rule P(A or B) = P(A) + P(B) − P(A and B) · testing for independence from numbers and tables · independent or mutually exclusive? · no calculator · exit quiz

Hook

14 · 15 · 120

In a large school, the probability that a student wears glasses is 14, the probability that a student is left-handed is 15, and the probability that a student is both is 120. Are wearing glasses and being left-handed independent?

Yes. In Year 1 you multiplied along the branches of a tree when two events were independent. Here 14 × 15 = 120, which is exactly P(both). So knowing that a student wears glasses tells you nothing about whether they are left-handed. If P(both) had been any other number, the events would not be independent. In this lesson you test events like this from the numbers alone.

Learn

Master the essentials

EventsThe rules
IndependentAND: P(A) × P(B)
OR: P(A) + P(B) − P(A)P(B)
Mutually exclusiveAND: 0
OR: P(A) + P(B)
Any two eventsAND: use the value given
OR: P(A) + P(B) − P(A and B)

The addition rule works for every pair of events. The other rows are special cases of it.

Independent events

One event does not change the probability of the other.

Two spins of the same fair spinner

The AND rule

For independent events only: P(A and B) = P(A) × P(B). It is also written P(A ∩ B).

P(A) = 0.2, P(B) = 0.9: 0.2 × 0.9 = 0.18

Testing for independence

A and B are independent exactly when P(A and B) is equal to P(A) × P(B).

13 × 34 = 14, so P(A and B) must be 14

Mutually exclusive

The events cannot happen together, so P(A and B) = 0.

One card: "a heart" and "a club"

The OR rule

Only for mutually exclusive events: P(A or B) = P(A) + P(B).

0.25 + 0.35 = 0.6

The addition rule

For ANY two events: P(A or B) = P(A) + P(B) − P(A and B). The overlap is counted once.

0.7 + 0.4 − 0.28 = 0.82

Neither

For independent events, P(neither) = P(not A) × P(not B).

P(A) = 0.8, P(B) = 0.5: 0.2 × 0.5 = 0.1

Not both

Two events with non-zero probabilities cannot be independent AND mutually exclusive: one rule gives 0 and the other does not.

P(A) = 0.2, P(B) = 0.3: 0.2 × 0.3 = 0.06, not 0

Worked examples

See it done

1. A and B are independent. P(A) = 0.4 and P(B) = 0.35. Find P(A and B) and P(A or B).

  1. They are independent, so multiply: P(A and B) = 0.4 × 0.35 = 0.14.
  2. Addition rule: P(A or B) = 0.4 + 0.35 − 0.14.
  3. = 0.75 − 0.14 = 0.61.
  4. Check: P(neither) = 0.6 × 0.65 = 0.39, and 1 − 0.39 = 0.61 ✓.

P(A and B) = 0.14; P(A or B) = 0.61

2. In a survey of 80 people, 48 drink coffee, 30 own a dog and 18 do both. Are drinking coffee and owning a dog independent?

  1. P(coffee) = 4880 = 35 and P(dog) = 3080 = 38.
  2. If they were independent, P(both) would be 35 × 38 = 940.
  3. The survey gives P(both) = 1880 = 940.
  4. The two values are equal, so the events are independent.

Yes, they are independent

3. A and B are independent. P(A) = 13 and P(A or B) = 35. Find P(B).

  1. Let P(B) = p. Then P(A and B) = 13p.
  2. Addition rule: 13 + p − 13p = 35.
  3. 23p = 35 − 13 = 415.
  4. p = 415 × 32 = 25.

P(B) = 25

Common mistake: Using P(A) × P(B) for any two events. Multiplying only works when the events are independent; if the question gives P(A and B), use that value. Also, "mutually exclusive" is not the same as "independent". Mutually exclusive events cannot happen together, so if one happens the chance of the other drops to 0. That is the opposite of independent.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Decide first: are the events independent, mutually exclusive, or neither? Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.A and B are independent. P(A) = 0.7 and P(B) = 0.3. Find P(A and B).

Not quite. This adds the probabilities. For "and" with independent events, multiply.

Solution

  1. The events are independent, so multiply.
  2. P(A and B) = 0.7 × 0.3 = 0.21.

Answer: D. 0.21.

Not quite. This is P(A or B) = 0.7 + 0.3 − 0.21.

Solution

  1. The events are independent, so multiply.
  2. P(A and B) = 0.7 × 0.3 = 0.21.

Answer: D. 0.21.

Not quite. This uses P(not A) = 0.3 instead of P(A) = 0.7.

Solution

  1. The events are independent, so multiply.
  2. P(A and B) = 0.7 × 0.3 = 0.21.

Answer: D. 0.21.

Correct.

Solution

  1. The events are independent, so multiply.
  2. P(A and B) = 0.7 × 0.3 = 0.21.

2.A fair six-sided dice is rolled once. A is the event "an even number" and B is the event "a 5". Find P(A or B).

Not quite. This takes away 12 × 16, as if the events were independent. One roll cannot be even AND a 5.

Solution

  1. A and B cannot happen together, so they are mutually exclusive.
  2. P(A or B) = 36 + 16 = 46.
  3. = 23.

Answer: B. 23.

Correct.

Solution

  1. A and B cannot happen together, so they are mutually exclusive.
  2. P(A or B) = 36 + 16 = 46.
  3. = 23.

Not quite. This multiplies. For "or", add.

Solution

  1. A and B cannot happen together, so they are mutually exclusive.
  2. P(A or B) = 36 + 16 = 46.
  3. = 23.

Answer: B. 23.

Not quite. This is P(A) only.

Solution

  1. A and B cannot happen together, so they are mutually exclusive.
  2. P(A or B) = 36 + 16 = 46.
  3. = 23.

Answer: B. 23.

3.P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.2. Are A and B independent?

Correct.

Solution

  1. Test: is P(A and B) equal to P(A) × P(B)?
  2. 0.5 × 0.4 = 0.2, which is P(A and B).
  3. So A and B are independent.

Not quite. Adding to 1 is not a test for independence.

Solution

  1. Test: is P(A and B) equal to P(A) × P(B)?
  2. 0.5 × 0.4 = 0.2, which is P(A and B).
  3. So A and B are independent.

Answer: A. Yes, because 0.5 × 0.4 = 0.2.

Not quite. That is the test for mutually exclusive events, not for independence.

Solution

  1. Test: is P(A and B) equal to P(A) × P(B)?
  2. 0.5 × 0.4 = 0.2, which is P(A and B).
  3. So A and B are independent.

Answer: A. Yes, because 0.5 × 0.4 = 0.2.

Not quite. That works out P(A or B). It does not test for independence.

Solution

  1. Test: is P(A and B) equal to P(A) × P(B)?
  2. 0.5 × 0.4 = 0.2, which is P(A and B).
  3. So A and B are independent.

Answer: A. Yes, because 0.5 × 0.4 = 0.2.

Exam standard (Several steps, like real exam questions.)

4.The two-way table shows whether 120 students play chess and whether they play an instrument. Which working shows that playing chess and playing an instrument are independent?

InstrumentNo instrumentTotal
Chess152540
No chess305080
Total4575120

120 students: do they play chess, and do they play an instrument?

Not quite. A sum less than 1 says nothing about independence.

Solution

  1. P(chess) = 40120 = 13 and P(instrument) = 45120 = 38.
  2. 13 × 38 = 18.
  3. P(both) = 15120 = 18. The values match, so the events are independent ✓.

Answer: B. 40120 × 45120 = 18 and 15120 = 18.

Correct.

Solution

  1. P(chess) = 40120 = 13 and P(instrument) = 45120 = 38.
  2. 13 × 38 = 18.
  3. P(both) = 15120 = 18. The values match, so the events are independent ✓.

Not quite. This multiplies the counts. Multiply the probabilities, then compare with P(both).

Solution

  1. P(chess) = 40120 = 13 and P(instrument) = 45120 = 38.
  2. 13 × 38 = 18.
  3. P(both) = 15120 = 18. The values match, so the events are independent ✓.

Answer: B. 40120 × 45120 = 18 and 15120 = 18.

Not quite. This shows the events are not mutually exclusive. It does not test independence.

Solution

  1. P(chess) = 40120 = 13 and P(instrument) = 45120 = 38.
  2. 13 × 38 = 18.
  3. P(both) = 15120 = 18. The values match, so the events are independent ✓.

Answer: B. 40120 × 45120 = 18 and 15120 = 18.

5.A and B are independent. P(A) = 0.3 and P(B) = 0.6. Find the probability that neither A nor B happens.

Correct.

Solution

  1. P(not A) = 1 − 0.3 = 0.7 and P(not B) = 1 − 0.6 = 0.4.
  2. The events are independent, so P(neither) = 0.7 × 0.4.
  3. = 0.28.

Not quite. This works out 1 − 0.3 − 0.6, as if the events were mutually exclusive.

Solution

  1. P(not A) = 1 − 0.3 = 0.7 and P(not B) = 1 − 0.6 = 0.4.
  2. The events are independent, so P(neither) = 0.7 × 0.4.
  3. = 0.28.

Answer: A. 0.28.

Not quite. This is P(A or B). Neither is 1 − P(A or B).

Solution

  1. P(not A) = 1 − 0.3 = 0.7 and P(not B) = 1 − 0.6 = 0.4.
  2. The events are independent, so P(neither) = 0.7 × 0.4.
  3. = 0.28.

Answer: A. 0.28.

Not quite. This is P(A and B).

Solution

  1. P(not A) = 1 − 0.3 = 0.7 and P(not B) = 1 − 0.6 = 0.4.
  2. The events are independent, so P(neither) = 0.7 × 0.4.
  3. = 0.28.

Answer: A. 0.28.

6.A and B are independent. P(B) = 37 and P(A and B) = 635. Find P(A or B).

Not quite. This adds P(A) and P(B) but does not take away P(A and B).

Solution

  1. Independent, so P(A) × 37 = 635.
  2. P(A) = 635 ÷ 37 = 635 × 73 = 25.
  3. P(A or B) = 1435 + 1535 − 635 = 2335.

Answer: C. 2335.

Not quite. This takes away P(A and B) twice.

Solution

  1. Independent, so P(A) × 37 = 635.
  2. P(A) = 635 ÷ 37 = 635 × 73 = 25.
  3. P(A or B) = 1435 + 1535 − 635 = 2335.

Answer: C. 2335.

Correct.

Solution

  1. Independent, so P(A) × 37 = 635.
  2. P(A) = 635 ÷ 37 = 635 × 73 = 25.
  3. P(A or B) = 1435 + 1535 − 635 = 2335.

Not quite. This is P(A). The question asks for P(A or B).

Solution

  1. Independent, so P(A) × 37 = 635.
  2. P(A) = 635 ÷ 37 = 635 × 73 = 25.
  3. P(A or B) = 1435 + 1535 − 635 = 2335.

Answer: C. 2335.

7.P(A) = 0.25, P(B) = 0.4 and P(A or B) = 0.65. Which statement is true?

Not quite. Both cannot be true: P(A and B) = 0, but 0.25 × 0.4 = 0.1.

Solution

  1. Addition rule: P(A and B) = P(A) + P(B) − P(A or B).
  2. = 0.25 + 0.4 − 0.65 = 0, so A and B are mutually exclusive.
  3. For independence, P(A and B) would be 0.25 × 0.4 = 0.1. It is 0, so they are not independent.

Answer: D. A and B are mutually exclusive, so they are not independent.

Not quite. P(A and B) = 0.25 + 0.4 − 0.65 = 0, not 0.1.

Solution

  1. Addition rule: P(A and B) = P(A) + P(B) − P(A or B).
  2. = 0.25 + 0.4 − 0.65 = 0, so A and B are mutually exclusive.
  3. For independence, P(A and B) would be 0.25 × 0.4 = 0.1. It is 0, so they are not independent.

Answer: D. A and B are mutually exclusive, so they are not independent.

Not quite. The addition rule gives P(A and B) from the numbers you have.

Solution

  1. Addition rule: P(A and B) = P(A) + P(B) − P(A or B).
  2. = 0.25 + 0.4 − 0.65 = 0, so A and B are mutually exclusive.
  3. For independence, P(A and B) would be 0.25 × 0.4 = 0.1. It is 0, so they are not independent.

Answer: D. A and B are mutually exclusive, so they are not independent.

Correct.

Solution

  1. Addition rule: P(A and B) = P(A) + P(B) − P(A or B).
  2. = 0.25 + 0.4 − 0.65 = 0, so A and B are mutually exclusive.
  3. For independence, P(A and B) would be 0.25 × 0.4 = 0.1. It is 0, so they are not independent.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A and B are independent. P(A) is twice P(B), and P(A and B) = 29. Find P(A or B).

Not quite. This adds P(A) and P(B) but does not take away P(A and B) = 29.

Solution

  1. Let P(B) = p, so P(A) = 2p.
  2. Independent: 2p × p = 29, so p² = 19 and p = 13.
  3. P(A) = 23 and P(B) = 13.
  4. P(A or B) = 23 + 13 − 29 = 79.

Answer: C. 79.

Not quite. This is P(B) only.

Solution

  1. Let P(B) = p, so P(A) = 2p.
  2. Independent: 2p × p = 29, so p² = 19 and p = 13.
  3. P(A) = 23 and P(B) = 13.
  4. P(A or B) = 23 + 13 − 29 = 79.

Answer: C. 79.

Correct.

Solution

  1. Let P(B) = p, so P(A) = 2p.
  2. Independent: 2p × p = 29, so p² = 19 and p = 13.
  3. P(A) = 23 and P(B) = 13.
  4. P(A or B) = 23 + 13 − 29 = 79.

Not quite. This takes away 29 twice.

Solution

  1. Let P(B) = p, so P(A) = 2p.
  2. Independent: 2p × p = 29, so p² = 19 and p = 13.
  3. P(A) = 23 and P(B) = 13.
  4. P(A or B) = 23 + 13 − 29 = 79.

Answer: C. 79.

9.A security system has a switch A and two alarms, B and C. It works if the switch works AND at least one alarm works. The parts work independently. P(A works) = 0.9, P(B works) = 0.6 and P(C works) = 0.5. Find the probability that the system works.

Correct.

Solution

  1. P(both alarms fail) = 0.4 × 0.5 = 0.2.
  2. P(at least one alarm works) = 1 − 0.2 = 0.8.
  3. The switch must also work: 0.9 × 0.8 = 0.72.

Not quite. This needs all three parts to work. Only one alarm is needed.

Solution

  1. P(both alarms fail) = 0.4 × 0.5 = 0.2.
  2. P(at least one alarm works) = 1 − 0.2 = 0.8.
  3. The switch must also work: 0.9 × 0.8 = 0.72.

Answer: A. 0.72.

Not quite. This is P(at least one alarm works). The switch must work too.

Solution

  1. P(both alarms fail) = 0.4 × 0.5 = 0.2.
  2. P(at least one alarm works) = 1 − 0.2 = 0.8.
  3. The switch must also work: 0.9 × 0.8 = 0.72.

Answer: A. 0.72.

Not quite. This multiplies 0.9 by P(both alarms FAIL) = 0.2.

Solution

  1. P(both alarms fail) = 0.4 × 0.5 = 0.2.
  2. P(at least one alarm works) = 1 − 0.2 = 0.8.
  3. The switch must also work: 0.9 × 0.8 = 0.72.

Answer: A. 0.72.

10.In a group of 60 people, 36 like tea and 25 like coffee. Liking tea and liking coffee are independent. How many of the 60 people like both?

Not quite. This multiplies the counts. Multiply the probabilities, then multiply by 60.

Solution

  1. P(tea) = 3660 = 35 and P(coffee) = 2560 = 512.
  2. Independent: P(both) = 35 × 512 = 14.
  3. 14 of 60 = 15 people.

Answer: B. 15.

Correct.

Solution

  1. P(tea) = 3660 = 35 and P(coffee) = 2560 = 512.
  2. Independent: P(both) = 35 × 512 = 14.
  3. 14 of 60 = 15 people.

Not quite. This works out 36 + 25 − 60, as if everyone liked at least one drink.

Solution

  1. P(tea) = 3660 = 35 and P(coffee) = 2560 = 512.
  2. Independent: P(both) = 35 × 512 = 14.
  3. 14 of 60 = 15 people.

Answer: B. 15.

Not quite. This is P(both). The question asks for the number of people.

Solution

  1. P(tea) = 3660 = 35 and P(coffee) = 2560 = 512.
  2. Independent: P(both) = 35 × 512 = 14.
  3. 14 of 60 = 15 people.

Answer: B. 15.

Apply

Test a claim

A shop asks 200 shoppers two questions. 120 brought their own bag, 70 paid by card, and 42 did both. The manager says, "Bringing a bag and paying by card are independent."

1. Find P(bag), P(card) and P(both) for a shopper chosen at random.

2. Is the manager right? Show your working.

3. Find the probability that a shopper chosen at random brought a bag or paid by card.

  1. P(bag) = 120200 = 35, P(card) = 70200 = 720, P(both) = 42200 = 21100.
  2. 35 × 720 = 21100, which equals P(both). Yes, the manager is right.
  3. Addition rule: 120200 + 70200 − 42200 = 148200 = 3750.

Next: the Lesson 1 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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