The rules for computing probabilities

Once the model is set, a few rules suffice to compute any probability. They can all be proved by counting outcomes.

The basic rules

0 ≤ P(A) ≤ 1                     a probability is never negative nor > 1
P(Ω) = 1                         something must happen
P(∅) = 0                         the impossible event

The complementary event

P(Ā) = 1 - P(A)

This is the most profitable rule of the chapter. Whenever a problem contains "at least one", think of it: the opposite of "at least one" is "none", almost always easier to count.

"Getting at least one 6 when rolling three dice"

  directly: at least one 6 = exactly one, or two, or three  -> 3 computations
  by the complement: no 6 -> (5/6)³ = 125/216

  P(at least one 6) = 1 - 125/216 = 91/216 ≈ 0.42

The union formula

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

We subtract P(A ∩ B) because the shared outcomes have been counted twice, once in each group.

   ______________________
  |    ___       ___     |
  |   / A \_____/ B \    |     the middle zone belongs to both:
  |  |    (#####)    |   |     counted twice in P(A) + P(B)
  |   \___/-----\___/    |
  |______________________|

When A and B are mutually exclusive (A ∩ B = ∅), the formula simplifies:

P(A ∪ B) = P(A) + P(B)

A numerical example: in a class, 50 % of pupils study German, 30 % Spanish, and 20 % study both.

P(German OR Spanish) = 0.50 + 0.30 - 0.20 = 0.60

It is not 0.80: the bilingual pupils count only once.

The method, in four steps

1. describe the sample space Ω and count its outcomes
2. check whether equiprobability is justified
3. translate the wording into events (∩, ∪, complement)
4. count the favourable outcomes, or go through the complement

Two classic traps

The two-way table. For two dice, it pays to display the 36 pairs:

        1    2    3    4    5    6
    1   2    3    4    5    6    7
    2   3    4    5    6    7    8
    3   4    5    6    7    8    9        each cell = 1 outcome out of 36
    4   5    6    7    8    9   10        (sum of the two dice)
    5   6    7    8    9   10   11
    6   7    8    9   10   11   12

One sees at once that the sum 7 appears 6 times (the diagonal), and the sum 12 only once.

Confusing "outcomes" with "values". There are 11 possible sums, but they are not equally likely: writing P(sum = 7) = 1/11 is wrong. It is the — famous — mistake d'Alembert himself made on an analogous problem.

Summary

• P(Ā) = 1 - P(A): the reflex for "at least one" problems.
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B): never count the intersection twice.
• If A and B are mutually exclusive, the formula reduces to P(A) + P(B).
• A two-way table makes the outcomes of a two-stage experiment visible.
• Never confuse the number of values with the number of outcomes.