Math Core

Probability is the language of chance, and it underlies everything in this unit: random samples, random assignment and margins of error all rely on it. A handful of rules let you find the probability of complicated events, like "A or B," "A and B" and "A given B," from simpler ones.

The basics

An experiment in probability is any process with uncertain results, like rolling a die. The set of all possible outcomes is the sample space, and an event is a set of outcomes. When all outcomes are equally likely,

P(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in the sample space}}.

Every probability is between $0$ (impossible) and $1$ (certain).

The complement of $A$ , written $A^c$ or "not $A$ ," is every outcome not in $A$ . Since $A$ and not $A$ together cover everything,

P(\text{not } A) = 1 - P(A).

The complement rule is especially useful for "at least one" questions. "At least one six in three rolls" has many cases, but its complement, "no sixes at all," has just one.

The addition rule: "A or B"

" $A$ or $B$ " means $A$ happens, $B$ happens, or both. If you add $P(A)$ and $P(B)$ , outcomes in both events get counted twice, so subtract them once.

The addition rule

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

If $A$ and $B$ are mutually exclusive (they can't both happen), then $P(A \text{ and } B) = 0$ and $P(A \text{ or } B) = P(A) + P(B)$ .

Worked example: Cards

One card is drawn from a standard $52$ -card deck. Find the probability that it is a king or a heart.

Solution. There are $4$ kings and $13$ hearts, and $1$ card (the king of hearts) is both:

P(\text{king or heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}.

Two-way tables and conditional probability

Sometimes you learn something that changes the probabilities. The probability of $B$ given that $A$ has happened is written $P(B \mid A)$ . Knowing $A$ happened shrinks the sample space to just the outcomes in $A$ .

Definition

Conditional probability

P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}.

With a table of counts, this is simply $\dfrac{\text{count in both } A \text{ and } B}{\text{count in } A}$ .

Worked example: Reading a two-way table

A survey of $200$ students asked whether they play a sport and whether they have a part-time job.

	Job	No job	Total
Sport	$30$	$70$	$100$
No sport	$20$	$80$	$100$
Total	$50$	$150$	$200$

A student is chosen at random. Find (1) $P(\text{job})$ , (2) $P(\text{sport or job})$ , (3) $P(\text{job} \mid \text{sport})$ and (4) $P(\text{sport} \mid \text{job})$ .

Solution.

$P(\text{job}) = \dfrac{50}{200} = 0.25$ .
$P(\text{sport or job}) = \dfrac{100}{200} + \dfrac{50}{200} - \dfrac{30}{200} = \dfrac{120}{200} = 0.6$ .
Look only at the Sport row: $P(\text{job} \mid \text{sport}) = \dfrac{30}{100} = 0.3$ .
Look only at the Job column: $P(\text{sport} \mid \text{job}) = \dfrac{30}{50} = 0.6$ .

Common mistake

$P(B \mid A)$ and $P(A \mid B)$ are usually different. In the example, $30\%$ of athletes have jobs, but $60\%$ of students with jobs are athletes. Always ask: which group is the "given" group? That group's total goes in the denominator.

Independence and the multiplication rule

Two events are independent if knowing that one happened doesn't change the probability of the other: $P(B \mid A) = P(B)$ . Coin flips and separate die rolls are independent. Rearranging the conditional probability formula gives the multiplication rule.

The multiplication rule

For any events,

P(A \text{ and } B) = P(A) \cdot P(B \mid A).

If $A$ and $B$ are independent, this becomes $P(A \text{ and } B) = P(A) \cdot P(B)$ . You can also use this as a test: $A$ and $B$ are independent exactly when $P(A \text{ and } B) = P(A) \cdot P(B)$ .

In the two-way table, $P(\text{job}) = 0.25$ but $P(\text{job} \mid \text{sport}) = 0.3$ . Knowing a student plays a sport changes the probability, so the events are not independent. Check with the product test: $P(\text{sport}) \cdot P(\text{job}) = 0.5 \times 0.25 = 0.125$ , but $P(\text{sport and job}) = \dfrac{30}{200} = 0.15$ .

Worked example: Drawing without replacement

A bag holds $5$ red and $3$ blue marbles. Two marbles are drawn one after the other without replacement. Find the probability that both are red.

Solution. The first draw is red with probability $\dfrac{5}{8}$ . Given that, $4$ red remain among $7$ marbles:

P(\text{both red}) = \frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}.

The draws are not independent, because the first draw changes what's left in the bag. With replacement, the answer would be $\dfrac{5}{8} \cdot \dfrac{5}{8} = \dfrac{25}{64}$ .

Worked example: At least one

A fair die is rolled $3$ times. Find the probability of at least one six.

Solution. The complement is "no sixes." Each roll is not a six with probability $\dfrac{5}{6}$ , and the rolls are independent:

P(\text{at least one six}) = 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421.

Tip

Don't confuse mutually exclusive with independent. Mutually exclusive events can't happen together, so if one happens the other's probability drops to $0$ . That means two mutually exclusive events (each with positive probability) are never independent.

Practice

Practice 1

The probability that it rains tomorrow is $0.35$ . What is the probability that it does not rain tomorrow?