Conditional Probability, SAT math lesson

Out of everyone at a school, only a few students are late on a given morning. Out of the students who ride the bus, the share who are late might be quite different. The question has changed even though the people have not, because “out of the students who ride the bus” picks a smaller group to look at. Conditional probability is the probability you get after a piece of information narrows the group, and SAT questions test whether you divide by that narrower group.

Key terms

Intersection

Definition The intersection of events $A$ and $B$ , written $A \cap B$ or ” $A$ and $B$ ”, is the event that both $A$ and $B$ occur. Its outcomes are those that belong to $A$ and to $B$ at the same time. In a two-way table, the intersection of a row category and a column category is the single inner cell where that row and column meet.

In plain words The intersection is the overlap, the members who fit both descriptions.

Example A school of $300$ students records who rides the bus and who was late at least once last month.

	Late	Not late	Total
Bus	$36$	$84$	$120$
No bus	$24$	$156$	$180$
Total	$60$	$240$	$300$

The event “rides the bus and was late” is the cell with $36$ students. For a student chosen at random from all $300$ ,

P(\text{bus and late}) = \frac{36}{300} = \frac{3}{25}.

Given

Definition In probability, the phrase “given $B$ ” means that the outcome is known to lie in the event $B$ , and only outcomes in $B$ are still possible. The probability of $A$ given $B$ is written $P(A \mid B)$ , where the event after the bar is the condition. In word problems the condition often appears as “if a student who rides the bus is chosen”, “among the students who were late” or “of those who”.

In plain words The word “given” tells you what you already know. Whatever follows “given”, “among” or “of those who” is the group you divide by.

Example “A student who rides the bus is chosen at random. What is the probability that the student was late?” The condition is riding the bus, so this asks for $P(\text{late} \mid \text{bus})$ . The question “A student who was late is chosen at random. What is the probability that the student rides the bus?” swaps the roles and asks for $P(\text{bus} \mid \text{late})$ .

Restricted sample space

Definition When the event $B$ is given, the restricted sample space is $B$ itself. The outcomes outside $B$ are removed, and, for equally likely outcomes or for counts in a table, each probability is recomputed as a fraction of the outcomes in $B$ .

In plain words Cover up every row or column that does not match the condition. What remains is the new whole, and its total is your denominator.

Example For $P(\text{late} \mid \text{bus})$ , keep only the bus row of the table. Its total is $120$ , and $36$ of those students were late, so

P(\text{late} \mid \text{bus}) = \frac{36}{120} = \frac{3}{10}.

The grand total $300$ plays no part, because the students who do not ride the bus are no longer possible.

Conditional probability

Definition For events $A$ and $B$ with $P(B) > 0$ , the conditional probability of $A$ given $B$ is

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

It is not defined when $P(B) = 0$ , since an impossible condition leaves nothing to divide by. With equally likely outcomes or a table of counts this is the same as

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

Rearranging gives the multiplication rule

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

In general $P(A \mid B)$ and $P(B \mid A)$ are different numbers, because they have the same numerator but different denominators.

In plain words Take the overlap and divide by the size of the condition. Reversing the condition changes the denominator, so it usually changes the answer.

Example From the bus table,

P(\text{late} \mid \text{bus}) = \frac{36}{120} = 0.3,

while

P(\text{bus} \mid \text{late}) = \frac{36}{60} = 0.6.

Thirty percent of bus riders were late, yet sixty percent of late students ride the bus. Both use the same $36$ students, but compare them with different groups. The formula agrees with the table, since $P(\text{bus and late}) = \frac{36}{300}$ and $P(\text{bus}) = \frac{120}{300}$ , and dividing gives $\frac{36}{120}$ again.

Independence

Definition Events $A$ and $B$ are independent when

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

When $P(B) > 0$ this is the same as $P(A \mid B) = P(A)$ , so knowing that $B$ happened does not change the probability of $A$ . Independence must be checked with actual probabilities or stated as an assumption, never guessed from wording that only sounds unrelated. It is also different from events that cannot happen together. If $P(A) > 0$ and $P(B) > 0$ and $A$ and $B$ have no outcomes in common, then $A$ and $B$ are not independent, because $P(A \text{ and } B) = 0$ while $P(A) \cdot P(B) > 0$ .

In plain words Independent means the condition makes no difference. Test it by comparing the conditional probability with the overall one.

Example In the bus table, $P(\text{late}) = \frac{60}{300} = 0.2$ but $P(\text{late} \mid \text{bus}) = 0.3$ , so being late and riding the bus are not independent. Compare a survey of $250$ adults.

	Owns a bicycle	No bicycle	Total
Left-handed	$10$	$15$	$25$
Right-handed	$90$	$135$	$225$
Total	$100$	$150$	$250$

Here $P(\text{bicycle}) = \frac{100}{250} = 0.4$ and $P(\text{bicycle} \mid \text{left-handed}) = \frac{10}{25} = 0.4$ . They are equal, so in this group owning a bicycle and being left-handed are independent. The product test agrees, since

\frac{25}{250} \cdot \frac{100}{250} = \frac{10}{250}.

Check yourself

A club survey counts juniors and seniors by the activity they chose. Of the juniors, $14$ chose chess and $26$ chose debate. Of the seniors, $21$ chose chess and $19$ chose debate. If a student who chose chess is selected at random, what is the probability that the student is a senior? (A) $\frac{21}{80}$ (B) $\frac{21}{40}$ (C) $\frac{7}{16}$ (D) $\frac{3}{5}$

Show the answer

The correct choice is (D). The condition is choosing chess, and $14 + 21 = 35$ students chose chess. Of them $21$ are seniors, so the probability is $\frac{21}{35} = \frac{3}{5}$ . Choice (A) divides by all $80$ students, which ignores the condition. Choice (B) divides by the $40$ seniors, which reverses the condition. Choice (C) equals $\frac{35}{80}$ , the probability that a randomly chosen student chose chess.

In a group of juniors and seniors who each chose art or music, $27$ seniors and $k$ juniors chose art. If a student who chose art is selected at random, the probability that the student is a junior is $\frac{2}{5}$ . What is the value of $k$ ?

Show the answer

The answer is $18$ . The reference group is the $k + 27$ students who chose art, so

\frac{k}{k + 27} = \frac{2}{5}.

Cross multiplying gives $5k = 2k + 54$ , so $k = 18$ . As a check, $\frac{18}{45} = \frac{2}{5}$ . Setting $\frac{k}{27} = \frac{2}{5}$ would use the seniors alone as the denominator, which is not the group described.

For two events, $P(A) = 0.5$ , $P(B) = 0.4$ and $P(A \text{ and } B) = 0.25$ . Which statement is true? (A) The events are independent, because $P(A \text{ and } B)$ is less than $P(A)$ . (B) The events are not independent, because $P(A) \cdot P(B) = 0.2$ , which is not $0.25$ . (C) The events are independent, because $P(A) + P(B)$ is less than $1$ . (D) The events are not independent, because $P(A \text{ and } B)$ is not $0$ .

Show the answer

The correct choice is (B). Independence requires $P(A \text{ and } B) = P(A) \cdot P(B)$ , and $0.5 \cdot 0.4 = 0.2 \ne 0.25$ . Equivalently, $P(A \mid B) = \frac{0.25}{0.4} = 0.625$ , which differs from $P(A) = 0.5$ . Choices (A) and (C) state facts that hold for many dependent events. Choice (D) confuses independent events with events that cannot happen together.

Conditional Probability

Theory, part 85

Practice, part 86

Key terms

Intersection

Given

Restricted sample space

Conditional probability

Independence

Check yourself

How this lesson connects

Builds on