Math Core

New information changes probabilities. The chance that a random roll of two number cubes has a sum of $12$ is $\dfrac{1}{36}$ . But if a friend peeks and tells you "at least one cube shows a $6$ ," the chance jumps. Conditional probability is the tool for updating a probability once you know something has happened.

Shrinking the sample space

Definition

Conditional probability

The conditional probability of $B$ given $A$ , written $P(B \mid A)$ , is the probability that $B$ happens when you already know that $A$ has happened. Read the bar as "given."

Knowing that $A$ happened throws away every outcome outside $A$ . The event $A$ becomes your new, smaller sample space, and you ask what fraction of it also lies in $B$ .

Worked example: Given a sum of 8

You roll two number cubes. Given that the sum is $8$ , what is the probability that one of the cubes shows a $3$ ?

The condition "sum is $8$ " leaves only these outcomes:

(2, 6), \quad (3, 5), \quad (4, 4), \quad (5, 3), \quad (6, 2).

That's the new sample space: $5$ equally likely outcomes. Two of them contain a $3$ : $(3, 5)$ and $(5, 3)$ . So

P(\text{a } 3 \mid \text{sum is } 8) = \frac{2}{5}.

Compare this with $P(\text{a } 3) = \dfrac{11}{36}$ with no information. Knowing the sum changed the probability.

The formula

Counting inside the smaller sample space works when outcomes are equally likely. In general, you divide the probability of the overlap by the probability of the condition.

Conditional probability formula

If $P(A) > 0$ , then

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

With equally likely outcomes this is the same as $\dfrac{\text{number of outcomes in } A \cap B}{\text{number of outcomes in } A}$ .

For the sum-of- $8$ example: $P(A \cap B) = \dfrac{2}{36}$ and $P(A) = \dfrac{5}{36}$ , so $P(B \mid A) = \dfrac{2/36}{5/36} = \dfrac{2}{5}$ , the same answer.

Common mistake

$P(B \mid A)$ and $P(A \mid B)$ are usually different. The probability that a person is a professional basketball player, given that they are over $7$ feet tall, is fairly high. The probability that a person is over $7$ feet tall, given that they are a professional basketball player, is much lower. The event after the bar is the one you know; it goes in the denominator.

The general multiplication rule

Multiply both sides of the formula by $P(A)$ and you get a rule for "and" that works even when events are dependent.

General multiplication rule

P(A \cap B) = P(A) \cdot P(B \mid A).

If $A$ and $B$ are independent, then $P(B \mid A) = P(B)$ and this becomes the rule from the last lesson, $P(A \cap B) = P(A) \cdot P(B)$ .

This is exactly what you need for drawing without replacement: the second probability is conditional on what happened in the first draw.

Worked example: Drawing without replacement

A bag has $4$ red and $6$ blue marbles. You draw two marbles without putting the first one back. What is the probability that both are red?

P(\text{1st red}) = \frac{4}{10}.

Given that the first was red, $3$ red marbles remain among $9$ :

P(\text{2nd red} \mid \text{1st red}) = \frac{3}{9}.

P(\text{both red}) = \frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15}.

With replacement, the answer would have been $\dfrac{4}{10} \cdot \dfrac{4}{10} = \dfrac{4}{25}$ , a bit larger.

Independence, again

The definition of independence ("knowing $A$ doesn't change the chance of $B$ ") now has a symbolic form.

Independence using conditional probability

$A$ and $B$ are independent exactly when $P(B \mid A) = P(B)$ (equivalently, $P(A \mid B) = P(A)$ ).

For example, if $P(B) = 0.3$ and $P(B \mid A) = 0.5$ , then learning that $A$ happened raised the chance of $B$ , so the events are dependent.

Tree diagrams

When an experiment happens in stages, draw a tree. Write the probability on each branch; branches after the first split are conditional probabilities. Multiply along a path to get the probability of that path, and add paths that lead to the same result.

Worked example: Two machines

A factory has two machines. Machine A makes $60\%$ of the parts, and $3\%$ of its parts are defective. Machine B makes the other $40\%$ , and $5\%$ of its parts are defective. A part is chosen at random.

What is the probability that it is defective?
If the part is defective, what is the probability that Machine A made it?

The tree has these four paths:

A, defective: $0.60 \times 0.03 = 0.018$
A, not defective: $0.60 \times 0.97 = 0.582$
B, defective: $0.40 \times 0.05 = 0.020$
B, not defective: $0.40 \times 0.95 = 0.380$

The four path probabilities add to $1$ , a good check.

Two paths end in "defective": $P(D) = 0.018 + 0.020 = 0.038$ .
Use the formula with "defective" as the condition:

P(A \mid D) = \frac{P(A \cap D)}{P(D)} = \frac{0.018}{0.038} = \frac{18}{38} = \frac{9}{19} \approx 0.47.

Even though Machine A makes most of the parts, a defective part is slightly more likely to have come from Machine B.

Tip

Before computing $P(B \mid A)$ , say the condition out loud: "Out of all the times $A$ happens, how often does $B$ happen?" Whatever follows "out of" goes in the denominator.