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The general addition rule

Probability · level 5, lesson 2 of 6

A question to start

In a class of 30 students, 18 play a sport and 15 play a musical instrument.

But \(18 + 15 = 33\), and there are only 30 students. Has somebody made a mistake?

No. Some students do both, and they were counted in both groups. If you are told that 8 students do both, you can work out exactly how many do at least one, and how many do neither. This lesson shows how, and turns it into a rule for any two events, whether they overlap or not.

The idea

"A and B": the overlap

In the last lesson, "A or B" meant A happens, or B happens, or both. Now we need a name for the "both" part. "A and B" is the event that \(A\) and \(B\) both happen: the outcomes that are in both groups. On a Venn diagram it is where the two loops overlap.

Take one card from an ordinary pack of 52. Let \(H\) be "the card is a heart" (13 cards) and \(K\) be "the card is a king" (4 cards). Exactly one card is both: the king of hearts. So "H and K" is just that one card.

121336 heartskingsboth neither

The four regions are: hearts that are not kings (12), the king of hearts (1), kings that are not hearts (3), and cards that are neither (36). Check: \(12 + 1 + 3 + 36 = 52\). ✓

The rule

How many cards are "heart or king"? If you add \(13 + 4 = 17\), the king of hearts has been counted twice: once as a heart and once as a king. Take it away once and you get \(17 - 1 = 16\). The diagram agrees: \(12 + 1 + 3 = 16\).

Dividing by 52 turns this into probabilities. That gives the general addition rule:

\[P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).\]

For the cards:

\[P(H \text{ or } K) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}.\]

Why the rule is true

Look at the three regions inside the loops in the picture. Call their sizes \(x\) (only \(A\)), \(y\) (both) and \(z\) (only \(B\)).

  • The loop \(A\) holds \(x + y\) outcomes.
  • The loop \(B\) holds \(y + z\) outcomes.
  • "A or B" holds \(x + y + z\) outcomes.

Adding the two loops gives \((x + y) + (y + z) = x + 2y + z\). That has one \(y\) too many. Subtracting the overlap \(y\) fixes it exactly:

\[(x + y) + (y + z) - y = x + y + z.\]

Every outcome in "A or B" is now counted exactly once. That is a complete proof when outcomes are equally likely. When they are not (a spinner, say), the same argument works with areas, or with the rule of lesson 1 applied to the three separate regions, which are mutually exclusive:

\[P(A \text{ or } B) = P(\text{only } A) + P(\text{both}) + P(\text{only } B).\]

This way of fixing a double count is called inclusion-exclusion: include everything, then exclude what was counted too often.

The old rule is a special case

If \(A\) and \(B\) are mutually exclusive, the overlap is empty, so \(P(A \text{ and } B) = 0\) and the rule becomes \(P(A) + P(B)\), exactly the rule of lesson 1. So you can always use the general rule. When the events cannot both happen, the last term is simply zero.

Using the rule backwards

The rule links four numbers: \(P(A)\), \(P(B)\), \(P(A \text{ or } B)\) and \(P(A \text{ and } B)\). Know any three and you can find the fourth. This is often how the rule is used in surveys, where people report "or" or "and" directly.

Some limits it gives

Because \(P(A \text{ and } B)\) is never negative, the rule shows that

\[P(A \text{ or } B) \le P(A) + P(B).\]

Adding chances can only over-estimate "or", never under-estimate it. And because "A or B" contains all of \(A\), \(P(A \text{ or } B)\) is at least as big as \(P(A)\) and at least as big as \(P(B)\).

Two-way tables do the bookkeeping

A two-way table sorts outcomes by two questions at once. For the class of 30 at the start, with 8 students who do both:

Instrument No instrument Total
Sport 8 10 18
No sport 7 5 12
Total 15 15 30

Fill it in from what you know: 8 in the top-left; then \(18 - 8 = 10\) across, \(15 - 8 = 7\) down; then the totals fix the last cell, \(12 - 7 = 5\). "Sport or instrument" is every cell except "no sport, no instrument": \(8 + 10 + 7 = 25\). The rule agrees: \(18 + 15 - 8 = 25\).

Three events (going further)

With three loops \(A\), \(B\), \(C\), adding the three counts counts the pairwise overlaps twice, so you take them away. But then the middle part, in all three loops, has been added 3 times and taken away 3 times, so it has vanished; add it back once:

\[P(A \text{ or } B \text{ or } C) = P(A) + P(B) + P(C) - P(A \text{ and } B) - P(A \text{ and } C) - P(B \text{ and } C) + P(A \text{ and } B \text{ and } C).\]

Example: how many whole numbers from 1 to 100 are divisible by 2, 3 or 5? Divisible by 2: 50. By 3: 33. By 5: 20. By both 2 and 3 (that is, by 6): 16. By 10: 10. By 15: 6. By all three (by 30): 3. So the count is \(50 + 33 + 20 - 16 - 10 - 6 + 3 = 74\), and a number picked at random from 1 to 100 has a \(\tfrac{74}{100}\) chance.

Worked examples

1. A die, overlapping events. One die is rolled. Find \(P(\text{even or more than 4})\).

  • Even \(= \{2, 4, 6\}\), so \(P = \tfrac36\). More than 4 \(= \{5, 6\}\), so \(P = \tfrac26\).
  • Both: \(\{6\}\), so \(P = \tfrac16\).
  • \(P(\text{even or more than 4}) = \tfrac36 + \tfrac26 - \tfrac16 = \tfrac46 = \tfrac23\).
  • Check by listing: \(\{2, 4, 5, 6\}\), 4 outcomes out of 6. ✓

2. The class. In the class of 30, find the probability that a student picked at random does at least one of the activities, and the probability they do neither.

  • \(P(\text{sport or instrument}) = \tfrac{18}{30} + \tfrac{15}{30} - \tfrac{8}{30} = \tfrac{25}{30} = \tfrac56\).
  • \(P(\text{neither}) = 1 - \tfrac56 = \tfrac16\), which is the 5 students in the table.

3. A survey in percentages. In a survey, 60% of people like pizza, 50% like burgers, and 30% like both. What percentage like at least one? What percentage like neither?

  • \(P(\text{pizza or burgers}) = 0.6 + 0.5 - 0.3 = 0.8\), so 80%.
  • Neither: \(1 - 0.8 = 0.2\), so 20%.

4. Using the rule backwards. \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \text{ or } B) = 0.7\). Find \(P(A \text{ and } B)\).

  • \(0.7 = 0.5 + 0.4 - P(A \text{ and } B)\).
  • So \(P(A \text{ and } B) = 0.9 - 0.7 = 0.2\).

5. Two dice. Two dice are rolled, one red and one blue. Find \(P(\text{red die shows 6, or total is at least 10})\).

  • Red shows 6: 6 outcomes, (6,1) to (6,6).
  • Total at least 10: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), which is 6 outcomes (red die first).
  • Both: (6,4), (6,5), (6,6), which is 3 outcomes.
  • \(P = \tfrac{6}{36} + \tfrac{6}{36} - \tfrac{3}{36} = \tfrac{9}{36} = \tfrac14\).

Try it

  1. A card is drawn from a full pack. Find \(P(\text{red or queen})\).
  2. One die is rolled. Find \(P(\text{odd or less than 3})\), and check by listing.
  3. \(P(A) = 0.45\), \(P(B) = 0.35\) and \(P(A \text{ and } B) = 0.15\). Find \(P(A \text{ or } B)\) and \(P(\text{neither})\).
  4. In a class of 32, 20 students have a dog, 14 have a cat, and 5 have neither. How many have both? Draw the Venn diagram with a number in each region.
  5. A card is drawn. Find \(P(\text{face card or spade})\). (The face cards are the jack, queen and king of each suit.)
  6. A whole number from 1 to 50 is picked at random. Find the probability it is a multiple of 3 or a multiple of 5.
  7. \(P(A) = 0.7\) and \(P(B) = 0.6\). What is the smallest possible value of \(P(A \text{ and } B)\)? The largest?
  8. A whole number from 1 to 60 is picked at random. Use the three-event rule to find the probability it is divisible by 2, 3 or 5.
Answers
  1. Red: 26 cards. Queens: 4. Both (red queens): 2. \(P = \dfrac{26 + 4 - 2}{52} = \dfrac{28}{52} = \dfrac{7}{13}\).
  2. Odd \(= \{1, 3, 5\}\), less than 3 \(= \{1, 2\}\), both \(= \{1\}\). \(P = \dfrac36 + \dfrac26 - \dfrac16 = \dfrac46 = \dfrac23\). Listing: \(\{1, 2, 3, 5\}\), 4 out of 6. ✓
  3. \(P(A \text{ or } B) = 0.45 + 0.35 - 0.15 = 0.65\). \(P(\text{neither}) = 1 - 0.65 = 0.35\).
  4. At least one pet: \(32 - 5 = 27\) students. So \(20 + 14 - \text{both} = 27\), and both \(= 34 - 27 = 7\). Regions: dog only \(20 - 7 = 13\), both 7, cat only \(14 - 7 = 7\), neither 5. Check: \(13 + 7 + 7 + 5 = 32\). ✓
  5. Face cards: 12. Spades: 13. Both (jack, queen, king of spades): 3. \(P = \dfrac{12 + 13 - 3}{52} = \dfrac{22}{52} = \dfrac{11}{26}\).
  6. Multiples of 3 up to 50: 16 (3 to 48). Of 5: 10. Of both, so of 15: 3 (15, 30, 45). Count \(= 16 + 10 - 3 = 23\), so \(P = \dfrac{23}{50}\).
  7. Since \(P(A \text{ or } B) \le 1\): \(0.7 + 0.6 - P(A \text{ and } B) \le 1\), so \(P(A \text{ and } B) \ge 0.3\). The smallest is 0.3. The overlap cannot be bigger than the smaller event, so the largest is 0.6 (when all of \(B\) lies inside \(A\)).
  8. Divisible by 2: 30. By 3: 20. By 5: 12. By 6: 10. By 10: 6. By 15: 4. By 30: 2. Count \(= 30 + 20 + 12 - 10 - 6 - 4 + 2 = 44\), so \(P = \dfrac{44}{60} = \dfrac{11}{15}\).

Common mistake

Forgetting to subtract the overlap. If your answer for "or" is bigger than it should be (or bigger than 1), you have probably counted the overlap twice. Always ask "what is in both?" before adding.

Subtracting when there is no overlap, or subtracting twice. For mutually exclusive events the overlap is empty, so nothing is taken away. And the overlap is taken away once, not twice: it was counted twice, and you want it counted once.

Where it leads