Experimental and theoretical probability
Toss a coin 1000 times and count the heads. The fraction you get, say 0.493, is an experimental probability, and it changes from one experiment to another. This chapter deals with theoretical (classical) probability, found by reasoning rather than by experiment.
Key terms
An experiment is an action with uncertain results: tossing a coin, throwing a die, drawing a card.
An outcome is one possible result.
Outcomes are equally likely if none is expected in preference to another. A fair coin's head and tail are equally likely.
An event is a collection of outcomes, such as "getting an even number".
The definition
When all outcomes are equally likely:
P(E) = number of outcomes favourable to E ÷ total number of possible outcomes
Example. A fair die is thrown. What is the probability of a number greater than 4?
Outcomes: 1, 2, 3, 4, 5, 6 (six in all). Favourable: 5, 6 (two). P = 2/6 = 1/3.
Important facts
0 ≤ P(E) ≤ 1 for every event.
An impossible event has probability 0: getting 7 on a single die.
A sure (certain) event has probability 1: getting a number less than 7 on a die.
An event with only one outcome is an elementary event. The probabilities of all elementary events of an experiment add up to 1.
Complementary events
The event "not E", written Ē, happens exactly when E doesn't. Since one of them must happen:
P(E) + P(not E) = 1
Example. The probability that it will rain tomorrow is 0.35. The probability that it will not rain is 1 − 0.35 = 0.65.
This is often the quickest way to answer "at least one" or "not" questions.
Coins
Two coins tossed together have 4 equally likely outcomes: HH, HT, TH, TT.
P(two heads) = 1/4
P(exactly one head) = 2/4 = 1/2 (HT and TH)
P(at least one head) = 3/4, or 1 − P(no heads) = 1 − 1/4
Three coins give 2³ = 8 outcomes.
Two dice
Two dice thrown together have 6 × 6 = 36 equally likely outcomes, written as pairs (first die, second die).
Sum | Pairs | Number |
|---|---|---|
2 | (1,1) | 1 |
7 | (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) | 6 |
12 | (6,6) | 1 |
P(sum 7) = 6/36 = 1/6; 7 is the most likely sum.
P(a doublet, the same number on both) = 6/36 = 1/6.
P(sum 13) = 0.
A pack of playing cards
A standard pack has 52 cards in 4 suits of 13 each: spades ♠ and clubs ♣ (black), hearts ♥ and diamonds ♦ (red). Each suit has an ace, numbers 2 to 10, and three face cards: king, queen and jack.
Event | Favourable cards | Probability |
|---|---|---|
a red card | 26 | 26/52 = 1/2 |
an ace | 4 | 4/52 = 1/13 |
a face card | 12 | 12/52 = 3/13 |
a red face card | 6 | 6/52 = 3/26 |
the queen of hearts | 1 | 1/52 |
not a diamond | 39 | 39/52 = 3/4 |
Bags of balls and other problems
Example. A bag contains 5 red, 8 white and 4 green balls. One ball is drawn at random. Total = 17.
P(red) = 5/17
P(not green) = 13/17
Example. A box contains 100 bulbs, of which 8 are defective. P(drawing a non-defective bulb) = 92/100 = 0.92.
Example. The probability of drawing a blue marble from a jar is 2/5. If the jar has 30 marbles, the number of blue marbles is 2/5 × 30 = 12.
Common mistakes
Treating outcomes as equally likely when they aren't. "Pass" and "fail" are two outcomes of an exam, but they are not equally likely.
Miscounting two-dice outcomes. (2, 5) and (5, 2) are different outcomes.
Forgetting that probability can't exceed 1 or be negative; such an answer signals an error.
Summary
P(E) = favourable outcomes ÷ total outcomes, for equally likely outcomes.
0 ≤ P(E) ≤ 1; impossible events have P = 0 and sure events P = 1.
P(E) + P(not E) = 1.
Two coins give 4 outcomes, two dice 36, and a pack of cards 52.