Elementary and Conditional Probability

Lessons

Senior 5 · Mathematics

Elementary and Conditional Probability

4 lessons

Lesson 3 of 4

Conditional probability and tree diagrams

45 min

When you already know something

A pupil is chosen at random from a school; the chance that she owns a bicycle is one number. If you are told that she is a day pupil, who travels to school every morning, the chance changes. Probability that takes known information into account is conditional probability. It is how a weather forecast updates after the morning clouds, and how a doctor reads a test result. When events affect each other, it replaces the simple multiplication rule of lesson 2.

Definition

The conditional probability of $A$ given $B$ , where $P (B) > 0$ , is

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

Knowing that $B$ occurred shrinks the sample space to $B$ ; with equally likely outcomes, $P(A \mid B) = \frac{n(A \cap B)}{n(B)}$ . Events are dependent when knowing one changes the probability of the other.

Worked example 1: from a two-way table

120 S5 pupils are classified as follows (illustrative):

	owns a bicycle	no bicycle	total
boarder	10	40	50
day pupil	35	35	70
total	45	75	120

A pupil is chosen at random. Find $P(\text{bicycle})$ , $P(\text{bicycle} \mid \text{day})$ and $P(\text{day} \mid \text{bicycle})$ .

Solution. $P(\text{bicycle}) = \frac{45}{120} = \frac{3}{8}$ . Given a day pupil, the sample space is the 70 day pupils: $P(\text{bicycle} \mid \text{day}) = \frac{35}{70} = \frac{1}{2}$ . Given a bicycle owner, it is the 45 owners: $P(\text{day} \mid \text{bicycle}) = \frac{35}{45} = \frac{7}{9}$ . Since $\frac{1}{2} \ne \frac{3}{8}$ , owning a bicycle and being a day pupil are dependent. Note that $P(A \mid B)$ and $P(B \mid A)$ are different numbers.

Important

Multiplication rule (general). Rearranging the definition,

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

Independence. When $P (B) > 0$ , $A$ and $B$ are independent if and only if $P(A \mid B) = P(A)$ : knowing $B$ does not change the chance of $A$ .

Properties of conditional probability. For a fixed event $B$ with $P (B) > 0$ , $P(\,\cdot \mid B)$ obeys all the rules of lesson 1:

$0 \le P(A \mid B) \le 1$ and $P(B \mid B) = P(\Omega \mid B) = 1$ ;
$P(A' \mid B) = 1 - P(A \mid B)$ ;
$P(A \cup C \mid B) = P(A \mid B) + P(C \mid B) - P(A \cap C \mid B)$ .

Reason for the complement. $A \cap B$ and $A' \cap B$ do not overlap and make up $B$ , so $P(A \cap B) + P(A' \cap B) = P(B)$ . Divide by $P (B)$ .

Tree diagrams

A tree diagram shows an experiment in stages. Each branch carries a conditional probability given the path so far, and the branches leaving one point add up to 1. The probability of a whole path is the product along it (the multiplication rule); the probability of an event made of several paths is the sum of their products (the addition rule for exclusive events).

Worked example 2: avocados without replacement

A basket holds 5 ripe and 3 unripe avocados. Two are taken at random, one after the other, without replacement. Find the probability that both are ripe, that exactly one is ripe, and that the first was ripe given that exactly one is ripe.

Solution. From the tree, $P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ . Exactly one ripe is the paths $R U$ and $U R$ : $\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$ . Then

P(\text{first ripe} \mid \text{exactly one ripe}) = \frac{P(RU)}{P(RU \cup UR)} = \frac{\frac{15}{56}}{\frac{30}{56}} = \frac{1}{2} .

Check: $20 + 15 + 15 + 6 = 56$ , so the four paths add to 1. ✓

Worked example 3: conditional probability from the rules

$P (A) = 0.6$ , $P (B) = 0.5$ and $P(A \cup B) = 0.8$ . Find $P(A \mid B)$ and $P(B \mid A)$ . Are $A$ and $B$ independent?

Solution. $P(A \cap B) = 0.6 + 0.5 - 0.8 = 0.3$ . So $P(A \mid B) = \frac{0.3}{0.5} = 0.6$ and $P(B \mid A) = \frac{0.3}{0.6} = 0.5$ . Since $P(A \mid B) = P(A)$ , they are independent.

Worked example 4: two dice, two different conditions

Two fair dice are rolled. Find $P(\text{total } 8 \mid \text{first die shows } 3)$ and $P(\text{first die shows } 3 \mid \text{total } 8)$ .

Solution. Given the first die is 3, the total is 8 only if the second shows 5: probability $\frac{1}{6}$ . A total of 8 has 5 outcomes, $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ , and one of them starts with 3: probability $\frac{1}{5}$ .

Warning

$P(A \mid B) \ne P(B \mid A)$ in general. Mixing them up is the commonest error in probability, in exams and in courtrooms.
Without replacement, the second-stage probabilities change: the denominators drop from 8 to 7.
Divide by $P (B)$ , the condition, not by $P (A)$ .

Tip

Read "given" as "among": $P(\text{bicycle} \mid \text{day})$ is "the fraction of bicycle owners among the day pupils". The word after "among" is the denominator.

Khan Academy: Conditional probability tree diagram example (about 11 min). Watch a two-stage tree built and then read backwards to find a conditional probability.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

$P (A) = 0.5$ , $P (B) = 0.4$ , $P(A \cap B) = 0.1$ . Find $P(A \mid B)$ and $P(B \mid A)$ .
Using the table of worked example 1, find $P(\text{boarder} \mid \text{no bicycle})$ .
A bag holds 4 red and 6 blue pens. Two are taken without replacement. Find the probability that both are blue, and that they are different colours.
On a given day it rains with probability 0.3 (illustrative). If it rains, a school bus is late with probability 0.4; if not, with probability 0.1. Draw a tree and find the probability that the bus is late.
Two fair dice are rolled and the total is 9. Find the probability that at least one die shows a 6.
Show that $P(A' \mid B) = 1 - P(A \mid B)$ for the data of question 1.
Prove that if $P(A \mid B) = P(A)$ , with $P (A) > 0$ and $P (B) > 0$ , then $P(B \mid A) = P(B)$ .

Show answersHide answers

$\frac{0.1}{0.4} = 0.25$ ; $\frac{0.1}{0.5} = 0.2$ .
$\frac{40}{75} = \frac{8}{15}$ .
$\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}$ ; $2 \times \frac{4}{10} \times \frac{6}{9} = \frac{8}{15}$ .
$0.3 \times 0.4 + 0.7 \times 0.1 = 0.12 + 0.07 = 0.19$ .
Totals of 9: $(3, 6), (4, 5), (5, 4), (6, 3)$ ; two contain a 6, so $\frac{2}{4} = \frac{1}{2}$ .
$P(A' \cap B) = P(B) - P(A \cap B) = 0.3$ , so $P(A' \mid B) = \frac{0.3}{0.4} = 0.75 = 1 - 0.25$ . ✓
$P(A \cap B) = P(B)P(A \mid B) = P(B)P(A)$ , so $P(B \mid A) = \frac{P(A \cap B)}{P(A)} = P(B)$ .

Sources and credits

Written by Ganzaa: the explanations, the tree diagram, the worked examples, the practice items and the quiz questions. The pupil, avocado and bus figures are illustrative. Every probability was checked by computer.
Video: Conditional probability tree diagram example | Probability | AP Statistics, Khan Academy, embedded from YouTube.

1When you already know something

2Tree diagrams

When you already know something

Tree diagrams