Unit 2 · Topic 2.4 · about 20 minutes

Introduction to Probability

List a sample space, find the probability of an event when outcomes are equally likely, and use the complement when "not E" is easier to count.

Predict first

You roll two fair dice. What is the probability that at least one of them shows a six?

The sample space

Every random process has a sample space: the set of all possible outcomes, listed so that no two of them overlap. For one die the sample space is {1, 2, 3, 4, 5, 6}. For a coin flipped twice it is {HH, HT, TH, TT}. Each trial produces exactly one outcome from the sample space, so the probability of the sample space as a whole is 1.

For two dice, the most useful sample space is the 36 ordered pairs (first die, second die). In the grid below each cell is one of those outcomes, and the number in it is the sum of the two dice.

Sums of two dice: rows are the first die, columns the second. All 36 cells are equally likely.
First die	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
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Equally likely outcomes

The probability of an event $E$ is written $P (E)$ . When all the outcomes in the sample space are equally likely, finding it comes down to counting:

P (E) = \frac{number of outcomes in E}{total number of outcomes in the sample space}

In the grid, the sum is 7 in six cells, so $P (sum = 7) = \frac{6}{36} = \frac{1}{6}$ . The sum is 2 in only one cell, so $P (sum = 2) = \frac{1}{36}$ .

Every probability is a number between 0 and 1, inclusive. An event with probability 0 never happens, like a sum of 13. An event with probability 1 always happens, like a sum between 2 and 12. If a calculation gives you 1.2 or a negative number, you have found an error, not an unusual event.

The complement rule

The complement of an event $E$ is the event that $E$ does not happen. It is written $E^{'}$ , $E$ or $E^{C}$ . Every outcome lands in $E$ or in its complement, never in both, so the two probabilities add to 1:

P (E^{C}) = 1 - P (E)

The complement earns its keep when "not $E$ " is easier to count than $E$ . "At least one six" fills the last row and the last column, and counting it directly means remembering not to count (6, 6) twice. Its complement, "no sixes", is the 5-by-5 block of cells where neither die shows a 6, which is 25 cells. So $P (at least one six) = 1 - \frac{25}{36} = \frac{11}{36}$ , the same answer the prediction reached by counting.

Worked exampleThree flips of a coin

A fair coin is flipped three times. List the sample space, then find the probability of exactly two heads and the probability of at least one tail.

List the sample space. Each flip has two possible results, so there are 8 equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Exactly two heads. Three outcomes qualify: HHT, HTH and THH. So $P (exactly two heads) = \frac{3}{8} = 0.375$ .
At least one tail, by counting. Every outcome except HHH contains a T, so seven of the eight qualify.
At least one tail, by the complement. The complement of "at least one tail" is "no tails", which is only HHH. So $P (at least one tail) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875$ .

Answer.

$P (exactly two heads) = 0.375$ and $P (at least one tail) = 0.875$ .

Lab

Probability Simulator

Roll two dice a few thousand times and compare the relative frequency of a sum of 7 with $\frac{6}{36} \approx 0.167$ . Then pick another event, work out its probability from the grid first, and see whether the simulation agrees.

Open the full Probability Simulator lab

Check your understanding

A bag holds 4 red, 5 blue and 3 green marbles. One marble is drawn at random. What is the probability that it is not blue?

A forecaster lists tomorrow's possible weather as sunny with probability 0.45, cloudy 0.30, rain 0.20 and snow 0.10. Exactly one of these will be the weather tomorrow. What is wrong with these probabilities?

Two fair dice are rolled. What is the probability that the sum is 10 or more? Give a decimal rounded to three places.

Your answer

A student says: "Flip two fair coins and there are three possible results: two heads, two tails, or one of each. So the probability of one of each is $\frac{1}{3}$ ." What is wrong with this reasoning?

For a randomly chosen student, which event is the complement of "the student has at least two siblings"?

Course alignment, for teachers

AP Statistics topic 2.4, Unit 2: Probability, Random Variables, and Probability Distributions.

Skill 3.C: Calculate and estimate expected counts, percentages, probabilities, and intervals.