The general addition rule
Two events can happen together. If the chance of each is known, why does adding them give too much for the chance that at least one happens, and by exactly how much?
Where you are 10 in Compound events
Why this exists
When two events cannot happen together, the chance that one or the other happens is the sum of their chances: that is the rule for mutually exclusive events. Most pairs of events are not like that. A zoo visitor can see the penguins and the reptile house on the same visit; a roll of two number cubes can show a 6 and have a large sum at the same time. Adding the two chances then gives the wrong answer, and sometimes an impossible one: two events with chances 0.6 and 0.7 would add to 1.3.
Recounting from scratch is not always possible either. A chance such as “55\% of visitors see the penguin pool” comes from records, not from a list of equally likely outcomes. What is needed is a rule that turns P(A), P(B) and one more number into P(A \text{ or } B), for any two events at all. The rule repairs the sum by removing what it counted twice, exactly as counting a union does for the sizes of sets.
Definition
Let A and B be two events on the same trial. In the language of unions and intersections of events:
- “A or B” is the event that at least one of A and B happens, possibly both. Its probability is written P(A \text{ or } B) or P(A \cup B).
- “A and B” is the event that both happen. Its probability is written P(A \text{ and } B) or P(A \cap B). This is the overlap.
The general addition rule says that for any two events A and B,
In words: add the two chances, then take away the chance of both once, because the sum counted it twice. Written with set signs it reads P(A \cup B) = P(A) + P(B) - P(A \cap B).
Intuition
Thirty lockers are numbered 1 to 30, and one is picked at random, so each has chance \displaystyle \frac{1}{30}. Let A be “the number is a multiple of 3” and B be “the number is a multiple of 5”. There are 10 multiples of 3 and 6 multiples of 5. Figure 1 sorts the lockers into a Venn diagram.
Lockers 15 and 30 are multiples of both, so they sit in the overlap. Adding 10 + 6 = 16 counts each of those two lockers once as a multiple of 3 and again as a multiple of 5. The lockers in either circle number
and dividing every count by 30 turns the counting statement into the probability statement:
The darker shading in Figure 1 is the picture of the whole idea. Each circle shades the overlap once, so the two together shade it twice. Subtracting the overlap once leaves every outcome in the union counted exactly once.
A Venn diagram can hold probabilities instead of counts. Then each of the four regions (only A, both, only B, neither) holds the chance of landing in that region and nowhere else. The regions share no outcome and together they hold every outcome, so their four probabilities add to 1.
Results
The general addition rule. For any two events A and B on the same trial,
The same rule, solved for the overlap. Any one of the four numbers can be found from the other three. In particular,
The regions of a Venn diagram.
Exactly one of the two. The chance that one happens but not the other is
The special case. If A and B are mutually exclusive, P(A \text{ and } B) = 0 and the rule becomes P(A \text{ or } B) = P(A) + P(B).
Limits on the overlap. The overlap cannot be larger than either event, and the union cannot be larger than 1. So
and the union lies between the larger of P(A), P(B) and the smaller of 1 and P(A) + P(B).
On a two-way table. When a table counts a group by two features, P(A) and P(B) come from a row total and a column total, which are the marginal probabilities, and P(A \text{ and } B) comes from the cell where that row and column cross, the joint probability.
Why it is true
Split “A or B” into three pieces: outcomes in A only, outcomes in both, and outcomes in B only. No outcome is in two of these pieces, so they are mutually exclusive, and by the addition rule for mutually exclusive events
The event A is made of two of the same pieces, “only A” and “both”, which also share nothing. So P(A) = P(\text{only } A) + P(A \text{ and } B), and in the same way P(B) = P(\text{only } B) + P(A \text{ and } B). Adding these two lines,
This is one copy of P(A \text{ and } B) more than the three-piece sum for P(A \text{ or } B). Removing that copy gives
The argument never needed the outcomes to be equally likely: it uses only the fact that the chance of an event made of separate pieces is the sum of the pieces’ chances. That is why the rule works for chances taken from records as well as for chances found by counting.
The other results follow from this one. The regions of the Venn diagram are the three pieces above plus “neither”, the complement of “A or B”. “Exactly one” is “only A” plus “only B”, which is P(A) + P(B) minus two copies of the overlap. For the limits, “A and B” is part of A, so its chance is at most P(A), and likewise at most P(B); and P(A \text{ or } B) \le 1 turns the rule into P(A) + P(B) - P(A \text{ and } B) \le 1, which rearranges to the lower limit.
Worked examples
Two zoo houses
At a zoo, a visitor chosen at random sees the penguin pool with probability 0.55, visits the reptile house with probability 0.40, and does both with probability 0.18. Find the probability that the visitor sees (a) at least one of the two, (b) neither, (c) exactly one. Then (d) fill in a Venn diagram of the four regions.
Solution(a) Seeing both is possible, so the events are not mutually exclusive and the general rule is needed. Let G be “sees the penguin pool” and R be “visits the reptile house”. Then
(b) “Neither” is the complement of “at least one”, so its chance is 1 - 0.77 = 0.23.
(c) The chance of the penguin pool only is 0.55 - 0.18 = 0.37, since the 0.55 includes the visitors who did both. The chance of the reptile house only is 0.40 - 0.18 = 0.22. These two pieces cannot overlap, so
The formula for exactly one agrees: 0.55 + 0.40 - 2(0.18) = 0.59.
(d) The four regions hold 0.37, 0.18, 0.22 and 0.23, as in Figure 2.
Check. 0.37 + 0.18 + 0.22 + 0.23 = 1, as the four regions must.
Reading the rule off a two-way table
A bike hire scheme records 160 hires by the kind of bike and the length of the trip.
| long trip | short trip | total | |
|---|---|---|---|
| electric | 54 | 38 | 92 |
| standard | 26 | 42 | 68 |
| total | 80 | 80 | 160 |
One hire is chosen at random. Find (a) P(\text{electric or long}) with the general rule, and check it by adding cells; (b) P(\text{standard or short}).
Solution(a) The chance of an electric bike is the row total over the grand total, \displaystyle \frac{92}{160}; the chance of a long trip is the column total, \displaystyle \frac{80}{160}. Hires that are both sit in the one cell where the electric row meets the long column: 54 of them. So
Check. The hires that are electric or long fill three cells: electric and long, electric and short, standard and long. That is 54 + 38 + 26 = 118. The fourth cell, standard and short, is the only one left out, and 160 - 42 = 118 as well. The 54 is in both the row total and the column total, which is why it is subtracted once.
(b) The standard row total is 68, the short column total is 80, and the cell where they cross holds 42:
Working back to the overlap
For an email arriving in an inbox, the chance that it has an attachment is 0.35, the chance that it comes from a colleague is 0.5, and the chance that it does at least one of these is 0.65. Find (a) the chance that it has an attachment and comes from a colleague, (b) the four regions of a Venn diagram, (c) the chance of exactly one of the two.
Solution(a) The general rule links the four numbers, and three are known. Let T be “has an attachment” and C be “from a colleague”. Solving for the overlap,
(b) Only an attachment: 0.35 - 0.2 = 0.15. Only from a colleague: 0.5 - 0.2 = 0.3. Neither: 1 - 0.65 = 0.35. With the overlap 0.2, the four regions add to 0.15 + 0.2 + 0.3 + 0.35 = 1.
(c) Exactly one is the two “only” regions: 0.15 + 0.3 = 0.45.
A large overlap with two number cubes
Two number cubes are rolled, and each of the 36 ordered pairs is equally likely. Find the probability that at least one cube shows a 1 or the sum is at most 4.
SolutionAt least one 1: there are 6 pairs with a 1 first and 6 with a 1 second, but (1,1) is in both lists, so 6 + 6 - 1 = 11 pairs. That count is itself the general rule at work.
A sum of at most 4: the sums 2, 3, 4 come from (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), so 6 pairs.
Both: every one of those six pairs except (2,2) contains a 1, so 5 pairs. Then
Check. The union is the 11 pairs with a 1 together with the one small-sum pair that has no 1, namely (2,2): 12 pairs. Here the overlap is most of the smaller event, so plain adding, \displaystyle \frac{17}{36}, would have been far too big.
When the overlap is not given
Events A and B have P(A) = 0.7 and P(B) = 0.5, and nothing else is known. Find the smallest and largest possible values of P(A \text{ and } B), and the range of P(A \text{ or } B).
SolutionThe overlap is part of B, so it is at most 0.5. That largest value happens when B lies entirely inside A.
The union is a probability, so it is at most 1. By the general rule,
which gives P(A \text{ and } B) \ge 0.2. That smallest value happens when A and B between them cover every outcome. So 0.2 \le P(A \text{ and } B) \le 0.5, and the two events cannot be mutually exclusive: their chances add to more than 1.
Putting the two ends of the overlap into the rule, P(A \text{ or } B) = 1.2 - P(A \text{ and } B) runs from 1.2 - 0.5 = 0.7 up to 1.2 - 0.2 = 1. Knowing P(A) and P(B) alone pins the union down only to this range; a third number, the overlap, is what fixes it.
Beyond two events
One of the 30 lockers of Figure 1 is chosen at random. Find the probability that its number is a multiple of 2, 3 or 5.
SolutionThe same repair works for three events, with one more step. Adding the three single counts, 15 multiples of 2, 10 of 3 and 6 of 5, counts a number lying in two of the sets twice. So subtract the three pairwise overlaps: multiples of 6 (5 of them), of 10 (3) and of 15 (2). Now look at locker 30, a multiple of all three: it was added three times and subtracted three times, so it is not counted at all, and it is added back once. The count is
so the probability is \displaystyle \frac{22}{30} = \frac{11}{15}.
Check. The numbers from 1 to 30 that are multiples of none of 2, 3, 5 are 1, 7, 11, 13, 17, 19, 23, 29: eight of them, and 30 - 8 = 22.
Boundaries
Both events belong to one trial. The rule asks whether a single outcome lies in A, in B, or in both, so A and B must be events in the same sample space. For two stages, like a roll and then a spin, the outcomes are the pairs, and both events have to be described in terms of pairs.
“Or” is inclusive. P(A \text{ or } B) includes the outcomes where both happen. If a problem means “one but not the other”, that is “exactly one”, and the overlap is subtracted twice, not once.
The rule needs the overlap. P(A) and P(B) alone do not fix P(A \text{ or } B), as the range found in Example 5 shows. The overlap has to be given, read from a table, counted, or found another way. One such way: for independent events, P(A \text{ and } B) = P(A) \times P(B). That product is only for independent events; for others the general multiplication rule is needed.
The numbers must fit together. Not every triple of numbers can be P(A), P(B) and P(A \text{ and } B). The overlap can be no larger than the smaller of P(A) and P(B) and no smaller than P(A) + P(B) - 1; outside these limits a Venn region would get a negative probability.
Three or more events need more terms. For three events, subtracting the three pairwise overlaps takes away too much from outcomes in all three, and the triple overlap is added back, as in Example 6. Each further event adds another layer of corrections.
Where this goes next
Rearranged, the rule computes the chance of “both” from the chance of “either”, and “both” is the starting point of Conditional probability: the chance of A given B is P(A \text{ and } B) \div P(B). The general multiplication rule runs the other way, building P(A \text{ and } B) from a conditional chance, and the two rules together handle any two events. Comparing P(A \text{ and } B) with P(A) \times P(B) is the test for independence.
Turning the question around, the chance that none of several events happens is often the quickest way to the chance that at least one does, which is the method of at least one problems. And the step from two events to three in Example 6 continues to any number of events, alternating adding and subtracting, which is the full inclusion-exclusion principle.
Exercises
Using the rule
- Events A and B have P(A) = 0.32, P(B) = 0.46 and P(A \text{ and } B) = 0.15. Find P(A \text{ or } B).
- A card is drawn at random from 40 cards numbered 1 to 40. Find the probability that its number is a multiple of 4 or a multiple of 6.
- A number cube is rolled once. Find the probability of an even number or a number greater than 3, with the general rule, and check by listing.
- Two number cubes are rolled. Find the probability that at least one cube shows a 3 or the sum is 7.
- A hotel guest chosen at random eats breakfast at the hotel with probability 0.62, uses the pool with probability 0.30, and does both with probability 0.21. Find the probability that the guest does at least one of these, and the probability that the guest does neither.
Two-way tables
A depot records 300 parcels by size and whether they arrived on time.
| on time | late | total | |
|---|---|---|---|
| small | 72 | 28 | 100 |
| medium | 90 | 30 | 120 |
| large | 48 | 32 | 80 |
| total | 210 | 90 | 300 |
One parcel is chosen at random.
- Find P(\text{small or on time}) with the general rule, and check by adding cells.
- Find P(\text{large or late}).
- Find P(\text{small or medium}), and say what the overlap is and why.
- Find P(\text{not medium or on time}) in two ways: with the general rule, and with the complement.
Venn diagrams and missing pieces
- P(A) = 0.5, P(B) = 0.35 and P(A \text{ or } B) = 0.7. Find P(A \text{ and } B), then the probabilities of the four regions of a Venn diagram.
- A Venn diagram shows 0.24 in “only A”, 0.16 in the overlap and 0.33 in “only B”. Find P(A), P(B), P(A \text{ or } B) and the probability of neither.
- P(A \text{ or } B) = 0.9, P(A) = 0.6 and P(A \text{ and } B) = 0.25. Find P(B).
- P(A) = 0.45, P(B) = 0.3 and P(A \text{ and } B) = 0.1. Find the probability that exactly one of A and B happens.
- Of 80 people asked, 46 read news online, 38 read news in print, and 15 do neither. One of the 80 is chosen at random. Find the probability that the person reads news both online and in print.
Limits and further cases
- P(A) = 0.8 and P(B) = 0.6. Find the smallest and largest possible values of P(A \text{ and } B), and the range of possible values of P(A \text{ or } B).
- Can events have P(A) = 0.3, P(B) = 0.4 and P(A \text{ and } B) = 0.35? Explain with a Venn diagram region.
- A and B are independent, with P(A) = 0.4 and P(B) = 0.25. Find P(A \text{ or } B) with the general rule, and check it through the chance that neither happens.
- A card is drawn at random from 42 cards numbered 1 to 42. Find the probability that its number is a multiple of 2, 3 or 7, using the three-event repair of Example 6.
Explaining
- Evander reads that the chance of rain on Saturday is 0.6 and on Sunday is 0.7, and concludes that the chance of rain at the weekend is 1.3. Explain what he has done wrong. If the chance of rain on both days is 0.45, find the chance of rain on at least one day.
- Explain why P(A \text{ or } B) \le P(A) + P(B) for any two events, and say exactly when the two sides are equal.
- Show that the chance of exactly one of A and B is P(A) + P(B) - 2P(A \text{ and } B).
- Explain why P(A \text{ and } B) \ge P(A) + P(B) - 1, and use it to show that two events with P(A) = 0.6 and P(B) = 0.7 cannot be mutually exclusive.
Answers
- 0.32 + 0.46 - 0.15 = 0.63.
- \displaystyle \frac{13}{40}. There are 10 multiples of 4 and 6 multiples of 6; the multiples of both are the multiples of 12, which are 12, 24, 36. So \displaystyle \frac{10 + 6 - 3}{40} = \frac{13}{40}.
- \displaystyle \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}, since \{2, 4, 6\} and \{4, 5, 6\} share 4 and 6. Listing: \{2, 4, 5, 6\}, four faces.
- At least one 3: 11 pairs. Sum 7: 6 pairs. Both: (3,4) and (4,3). So \displaystyle \frac{11 + 6 - 2}{36} = \frac{15}{36} = \frac{5}{12}.
- At least one: 0.62 + 0.30 - 0.21 = 0.71. Neither: 1 - 0.71 = 0.29.
- \displaystyle \frac{100 + 210 - 72}{300} = \frac{238}{300} = \frac{119}{150}. Cells: 72 + 28 + 90 + 48 = 238.
- \displaystyle \frac{80 + 90 - 32}{300} = \frac{138}{300} = \frac{23}{50}.
- \displaystyle \frac{100 + 120}{300} = \frac{220}{300} = \frac{11}{15}. The overlap is 0: a parcel has only one size, so the events are mutually exclusive.
- “Not medium” holds 100 + 80 = 180 parcels, and “not medium and on time” holds 72 + 48 = 120. So \displaystyle \frac{180 + 210 - 120}{300} = \frac{270}{300} = \frac{9}{10}. Complement: the only parcels left out are medium and late, so \displaystyle 1 - \frac{30}{300} = \frac{9}{10}.
- P(A \text{ and } B) = 0.5 + 0.35 - 0.7 = 0.15. Only A: 0.35. Only B: 0.2. Neither: 0.3. Check: 0.35 + 0.15 + 0.2 + 0.3 = 1.
- P(A) = 0.24 + 0.16 = 0.40, P(B) = 0.33 + 0.16 = 0.49, P(A \text{ or } B) = 0.24 + 0.16 + 0.33 = 0.73, neither 1 - 0.73 = 0.27.
- 0.9 = 0.6 + P(B) - 0.25, so P(B) = 0.55.
- 0.45 + 0.3 - 2(0.1) = 0.55, the sum of only A, 0.35, and only B, 0.2.
- \displaystyle \frac{19}{80}. At least one kind: 80 - 15 = 65 people. Both: 46 + 38 - 65 = 19.
- The overlap is at most 0.6 (it is part of B) and at least 0.8 + 0.6 - 1 = 0.4. So 0.4 \le P(A \text{ and } B) \le 0.6, and P(A \text{ or } B) = 1.4 - P(A \text{ and } B) runs from 0.8 to 1.
- No. The region “only A” would hold 0.3 - 0.35 = -0.05, and no probability is negative. The overlap is part of A, so it cannot exceed 0.3.
- Independent, so P(A \text{ and } B) = 0.4 \times 0.25 = 0.1, and P(A \text{ or } B) = 0.4 + 0.25 - 0.1 = 0.55. Check: neither has chance 0.6 \times 0.75 = 0.45, and 1 - 0.45 = 0.55.
- \displaystyle \frac{5}{7}. Multiples of 2, 3, 7: 21, 14, 6. Of 6, 14, 21: 7, 3, 2. Of 42: 1. So 21 + 14 + 6 - 7 - 3 - 2 + 1 = 30, and \displaystyle \frac{30}{42} = \frac{5}{7}.
- Rain on both days is possible, so the events are not mutually exclusive and adding counts a wet weekend twice; a probability above 1 is the sign. With the general rule, 0.6 + 0.7 - 0.45 = 0.85.
- P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B), and P(A \text{ and } B) \ge 0, so the union is at most P(A) + P(B). The sides are equal exactly when P(A \text{ and } B) = 0, which for the experiments here means the events are mutually exclusive.
- Exactly one is “only A” or “only B”, which share nothing, so its chance is (P(A) - P(A \text{ and } B)) + (P(B) - P(A \text{ and } B)), which is P(A) + P(B) - 2P(A \text{ and } B).
- P(A \text{ or } B) \le 1, so P(A) + P(B) - P(A \text{ and } B) \le 1, which rearranges to P(A \text{ and } B) \ge P(A) + P(B) - 1. With 0.6 and 0.7 the overlap is at least 0.3, so it is not 0 and the events can happen together.