4.6 Probability

1. Sample Space and Events

We start again with a universal set $S$. In probability theory, this set is known as the sample space; it contains all possible outcomes of a game, experiment, etc. The subsets $A, B, \dots$ of the sample space $S$ are called events.

Consider the sample space $S$. The number of elements in $S$, that is $n(S)$, is denoted by $\text{TOTAL}$. The probability of some event $A$ is simply defined by:

$P(A) = \dfrac{n(A)}{\text{TOTAL}}$

For example, in the following Venn diagram, the sample space $S$ contains 100 elements, while the event $A$ contains 30 elements.

$P(A) = \dfrac{n(A)}{\text{TOTAL}} = \dfrac{30}{100} = 0.3$

In simple words, if we choose an element from $S$ at random (provided that every element is equally likely to be selected), the probability that this element belongs to $A$ is 30 out of 100, otherwise 30% (that is $0.3$).

Clearly, it holds that: $0 \le P(A) \le 1$, $P(\emptyset) = 0$, and $P(S) = 1$.

2. Complementary and Combined Events

Complementary Events: In our example above, $P(A') = 0.7$. In general:

$P(A') = 1 - P(A)$

Combined Events: Remember the basic property for combined events: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. If we divide all terms by the $\text{TOTAL}$, we obtain:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For example, consider the Venn diagram (left) mapping the number of elements.
We can find the respective probabilities:
$P(A) = 0.3$, $\quad P(B) = 0.4$
$P(A') = 0.7$, $\quad P(B') = 0.6$
$P(A \cap B) = 0.1$, $\quad P(A \cup B) = 0.6$
Also, applying the formula confirms:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.6 = 0.3 + 0.4 - 0.1$

(Diagram containing probabilities)

3. Mutually Exclusive Sets

We have seen that two events are mutually exclusive if $A \cap B = \emptyset$ or equivalently $n(A \cap B) = 0$. Equivalently if:

$P(A \cap B) = 0$

In this case only, the addition rule simplifies to:

$P(A \cup B) = P(A) + P(B)$

EXAMPLE 1

Given that $P(A) = 0.5$, $P(B) = 0.3$, and $P(A \cup B) = 0.6$, let us construct a Venn diagram representing the combined events A and B.

Notice that $P(A \cup B) \ne P(A) + P(B)$ ($0.6 \ne 0.8$). The difference implies the existence of an intersection:

$P(A \cap B) = 0.2$

Starting from the intersection $0.2$, we may easily complete the Venn diagram.

After completing the diagram, we can answer any probability question. For example:

$P(A \cap B') = 0.3$
$P(A \cup B') = 0.9$
$P(A' \cap B) = 0.1$
$P(A' \cup B) = 0.7$
$P(A' \cap B') = 0.4$
$P(A' \cup B') = 0.8$

4. Probability and Tables

Another way to represent sets in order to find probabilities is tabular form. It is appropriate when the sample space is partitioned into disjoint subsets according to two different criteria; for example, MALE-FEMALE and SMOKERS-NON SMOKERS.

EXAMPLE 2

Let us consider the following group of 200 people:

	Male	Female	Total
Smoker	40	20	60
Non-smoker	80	60	140
Total	120	80	200

In order to find the probability of a group (or combination of groups) we simply divide its size by 200, the total number of people. If we select a person at random, the probability that this person is:

Male is: $P(\text{male}) = \dfrac{120}{200} = 0.6$
Female is: $P(\text{female}) = \dfrac{80}{200} = 0.4$
Smoker is: $P(\text{smoker}) = \dfrac{60}{200} = 0.3$
Non-smoker is: $P(\text{non-smoker}) = \dfrac{140}{200} = 0.7$
Male AND Smoker: $P(\text{male} \cap \text{smoker}) = \dfrac{40}{200} = 0.2$
Male OR Smoker: $P(\text{male} \cup \text{smoker}) = \dfrac{140}{200} = 0.7$

Notice: In the last probability, we consider the column of male and the row of smokers, but the combination male-smoker is counted only once. It holds again:
$P(\text{male} \cup \text{smoker}) = P(\text{male}) + P(\text{smoker}) - P(\text{male} \cap \text{smoker})$.

5. Counting Outcomes: Two Dice

Some problems require particular techniques for counting the appropriate group size. Tossing two dice is a characteristic example.

EXAMPLE 3

We toss two dice. There are 36 possible outcomes (combinations of scores). The following table helps to visualize the outcomes:

Notice that there is only one combination for ones (the first dot; 1-1) but two combinations of one-two (1-2 and 2-1). We find the following probabilities:

$P(\text{two sixes}) = \dfrac{1}{36}$ (the very last dot)
$P(\text{at least one six}) = \dfrac{11}{36}$ (last column and last row)
$P(\text{exactly one six}) = \dfrac{10}{36}$ (removes the 6-6 corner dot)
$P(\text{same score}) = \dfrac{6}{36}$ (the main diagonal: 1-1, etc.)
$P(\text{sum of scores} = 9) = \dfrac{4}{36}$ (the red dotted line)
$P(\text{sum of scores} > 9) = \dfrac{6}{36}$ (below the dotted line)
$P(\text{sum of scores} < 9) = \dfrac{26}{36}$ (above the dotted line)