IB Analysis and Approaches HL · Probability · Section 3 of 5

Probability and its properties: Laplace, complement and union

3 worked examples · By Álvaro López

Key idea

Probability is a number from $0$ to $1$ that measures how likely an event is. With Laplace's rule, the complement and the union formula, and using a Venn diagram or a table, you solve most of the basic probability questions in the IB exam, often without a calculator.

What probability measures

If you repeat an experiment $N$ times and the event $S$ occurs $f (S)$ times (absolute frequency), the relative frequency is $fr (S) = \frac{f ( S )}{N}$ As $N$ increases, the relative frequency gets closer to a fixed value, which is the probability $P (S)$ . This stabilisation is called the law of large numbers. That is why a relative frequency measured with lots of data can be used as an estimate of the probability. With few repetitions it can be quite far off: 3 heads in 5 tosses does not prove that the coin is biased.

Relative frequencies are always numbers between $0$ and $1$ ; that of the certain event is $1$ ; and if two events cannot occur at the same time, the relative frequency of their union is the sum of the two. These three ideas are the starting rules of probability.

$P (A) = 0$ indicates an event that never occurs and $P (A) = 1$ one that always occurs. A probability outside $[0, 1]$ , or a negative one, is a sign of an error.

Laplace's rule

Each elementary outcome has weight $\frac{1}{n ( U )}$ and $A$ contains $n (A)$ of them. It can only be used if the outcomes are equally likely: a coin, a die, or a ball drawn "at random" from an urn all satisfy this; a loaded die does not. Cases are counted by listing them, using a table of outcomes (each cell is an equally likely outcome) or with the combinatorics from the combinatorics section of this topic.

Properties

Why they hold:

Complement. $A$ and $A^{'}$ are mutually exclusive and together make up $U$ , so $P (A) + P (A^{'}) = P (U) = 1$ . This is the key property: for "at least one" or "ever", calculate the opposite and subtract from $1$ .
Union. When you add $P (A) + P (B)$ the common region is counted twice; subtract it once.
Inclusion. If $A$ lies inside $B$ , then $B$ contains everything in $A$ and possibly something more.

Rearranging the union formula gives a very useful way of finding the intersection: $P (A \cap B) = P (A) + P (B) - P (A \cup B)$

The four regions and their probabilities

Events $A$ and $B$ divide $U$ into four disjoint regions, whose probabilities add up to $1$ : $P (A \cap B) + P (A \cap B^{'}) + P (A^{'} \cap B) + P (A^{'} \cap B^{'}) = 1$ Three very useful relationships follow from this: $P (A) = P (A \cap B) + P (A \cap B^{'}) P (B) = P (A \cap B) + P (A^{'} \cap B) P (A \cup B) = 1 - P (A^{'} \cap B^{'})$ The third is De Morgan's law in probability form: anything that is not outside both circles lies in the union.

Venn diagram with U and two overlapping circles A and B, with the four regions in different colours and labelled: only A (A ∩ B′) in green, the common region (A ∩ B) in blue, only B (A′ ∩ B) in orange and the outside (A′ ∩ B′) in grey; on the right, P(A) = P(A ∩ B) + P(A ∩ B′), P(B) = P(A ∩ B) + P(A′ ∩ B) and the four regions add up to 1

Contingency tables

A contingency table sorts outcomes by two criteria and is a Venn diagram in disguise: each inner cell is an intersection and the totals are the probabilities of each event. For example, in a group people are classified according to whether they cycle ( $A$ ) or not, and whether they live in the city ( $B$ ) or not:

	$B$	$B^{'}$	Total
$A$	$P (A \cap B)$	$P (A \cap B^{'})$	$P (A)$
$A^{'}$	$P (A^{'} \cap B)$	$P (A^{'} \cap B^{'})$	$P (A^{'})$
Total	$P (B)$	$P (B^{'})$	$1$

Each row and each column adds up to its total, and the grand total is $1$ (or $n (U)$ , if the table shows frequencies).

Worked examples

Example 1 Basic

Coloured balls

A bag contains 20 balls of the same size: 8 red, 7 blue and 5 green. One ball is drawn at random. Calculate the probability that

a) it is red;

b) it is not green;

c) it is red or blue;

d) it is neither blue nor green.

Solution

All balls are equally likely and $n (U) = 20$ .

a) $P (red) = \frac{8}{20} = \frac{2}{5}$ .

b) By the complement, $P (not green) = 1 - \frac{5}{20} = \frac{3}{4}$ .

c) Red and blue are mutually exclusive, so they are added: $\frac{8}{20} + \frac{7}{20} = \frac{15}{20} = \frac{3}{4}$ . (This agrees with part b, because "red or blue" is the same as "not green".)

d) "Neither blue nor green" is "red": $\frac{8}{20} = \frac{2}{5}$ .

Answer: a) $\frac{2}{5}$ ; b) $\frac{3}{4}$ ; c) $\frac{3}{4}$ ; d) $\frac{2}{5}$ .

Example 2 Medium

Sum of two dice

Two fair 6-sided dice are rolled and the results are added. Calculate the probability that the sum is

a) 9;

b) at least 10;

c) even;

d) even or a multiple of 5.

Solution

The table of outcomes has $6 \cdot 6 = 36$ equally likely cells.

a) The sum 9 is obtained with $(3, 6), (4, 5), (5, 4), (6, 3)$ : 4 cases. $P = \frac{4}{36} = \frac{1}{9}$ .

b) "At least 10" groups together the sums 10 (3 cases), 11 (2 cases) and 12 (1 case): 6 cases. $P = \frac{6}{36} = \frac{1}{6}$ .

c) The sum is even when both dice have the same parity: $3 \cdot 3 + 3 \cdot 3 = 18$ cases. $P = \frac{18}{36} = \frac{1}{2}$ .

d) Let $A$ : "even" and $B$ : "multiple of 5". The multiples of 5 are the sums 5 (4 cases) and 10 (3 cases), so $n (B) = 7$ . Of these, only the sum 10 is even: $n (A \cap B) = 3$ . By the union formula, $n (A \cup B) = 18 + 7 - 3 = 22 ⟹ P (A \cup B) = \frac{22}{36} = \frac{11}{18}$

Answer: a) $\frac{1}{9}$ ; b) $\frac{1}{6}$ ; c) $\frac{1}{2}$ ; d) $\frac{11}{18}$ .

Example 3 Hard

Union probabilities and bounds

For two events $A$ and $B$ it is known that $P (A) = \frac{3}{5}$ , $P (B) = \frac{1}{2}$ and $P (A^{'} \cap B^{'}) = \frac{1}{10}$ .

a) Find $P (A \cup B)$ .

b) Find $P (A \cap B)$ .

c) Find $P (A \cap B^{'})$ .

d) Find $P (A^{'} \cup B^{'})$ .

e) Without using the data $P (A^{'} \cap B^{'})$ , what is the smallest value that $P (A \cap B)$ can take, given $P (A) = \frac{3}{5}$ and $P (B) = \frac{1}{2}$ ?

Solution

a) What lies outside the union is $A^{'} \cap B^{'}$ (De Morgan), so $P (A \cup B) = 1 - P (A^{'} \cap B^{'}) = 1 - \frac{1}{10} = \frac{9}{10}$

b) From the union formula, $P (A \cap B) = P (A) + P (B) - P (A \cup B) = \frac{3}{5} + \frac{1}{2} - \frac{9}{10} = \frac{6 + 5 - 9}{10} = \frac{1}{5}$

c) "Only $A$ ": $P (A \cap B^{'}) = P (A) - P (A \cap B) = \frac{3}{5} - \frac{1}{5} = \frac{2}{5}$ .

d) $A^{'} \cup B^{'} = (A \cap B)^{'}$ , so $P (A^{'} \cup B^{'}) = 1 - \frac{1}{5} = \frac{4}{5}$ .

e) As $P (A \cup B) \leq 1$ , from $P (A \cap B) = P (A) + P (B) - P (A \cup B)$ we get $P (A \cap B) \geq P (A) + P (B) - 1 = \frac{3}{5} + \frac{1}{2} - 1 = \frac{1}{10}$ The minimum is $\frac{1}{10}$ , and it occurs when $A \cup B = U$ (nothing is left outside the circles). Moreover the circles must overlap: with these probabilities $A$ and $B$ cannot be mutually exclusive.

In the exam: name the property you use at each step ("complement", "union") and redraw the diagram with the four values obtained, which must add up to $1$ : $\frac{1}{5} + \frac{2}{5} + \frac{3}{10} + \frac{1}{10} = 1$ . If there is an arithmetic slip, an answer without working loses the method marks.

Common mistakes

Summary

Relative frequency $fr (S) = \frac{f ( S )}{N}$ ; with many repetitions it stabilises at $P (S)$ (law of large numbers).
Axioms: $0 \leq P (A) \leq 1$ , $P (U) = 1$ and, if $A \cap B = \emptyset$ , $P (A \cup B) = P (A) + P (B)$ .
Laplace (equally likely outcomes): $P (A) = \frac{n ( A )}{n ( U )}$ .
$P (A^{'}) = 1 - P (A)$ ; $P (\emptyset) = 0$ ; $A \subset B \Rightarrow P (A) \leq P (B)$ .
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , and from it $P (A \cap B) = P (A) + P (B) - P (A \cup B)$ ; $P (A \cup B) = 1 - P (A^{'} \cap B^{'})$ .
The four regions of a Venn diagram add up to $1$ ; a contingency table is the same idea with rows and columns.
For "at least one", calculate the complement and subtract from $1$ .

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