The addition law

1 The addition law

As we have already noted, the sample space $S$ is the set of all possible outcomes of a given experiment. Certain events $A$ and $B$ are subsets of $S$ . In the previous Section we defined what was meant by $P (A), P (B)$ and their complements in the particular case in which the experiment had equally likely outcomes.

Events, like sets, can be combined to produce new events.

$A \cup B$ denotes the event that event $A$ or event $B$ (or both) occur when the experiment is performed.
$A \cap B$ denotes the event that both $A$ and $B$ occur together.

In this Section we obtain expressions for determining the probabilities of these combined events, which are written $P (A \cup B)$ and $P (A \cap B)$ respectively.

1.1 Types of events

There are two types of events you will need to able to identify and work with: mutually exclusive events and independent events. (We deal with independent events in subsection 3.)

1.2 Mutually exclusive events

Mutually exclusive events are events that by definition cannot happen together. For example, when tossing a coin, the events ‘head’ and ‘tail’ are mutually exclusive; when testing a switch ‘operate’ and ‘fail’ are mutually exclusive; and when testing the tensile strength of a piece of wire, ‘hold’ and ‘snap’ are mutually exclusive. In such cases, the probability of both events occurring together must be zero. Hence, using the usual set theory notation for events $A$ and $B$ , we may write:

$P (A \cap B) = 0, provided that A and B are mutually exclusive events$

Task!

Decide which of the following pairs of events ( $A$ and $B$ ) arising from the experiments described are mutually exclusive.

Two cards are drawn from a pack
$A = {a red card is drawn}$

$B = {a picture card is drawn}$
The daily traffic accidents in Loughborough involving pedal cyclists and motor cyclists are counted
$A = {three motor cyclists are injured in collisions with cars}$

$B = {one pedal cyclist is injured when hit by a bus}$
A box contains 20 nuts. Some have a metric thread, some have a British Standard Fine (BSF) thread and some have a British Standard Whitworth (BSW) thread.
$A = {first nut picked out of the box is BSF}$

$B = {second nut picked out of the box is metric}$

$A$ and $B$ are not mutually exclusive.
$A$ and $B$ are mutually exclusive.
$A$ and $B$ are not mutually exclusive.

Key Point 5

The Addition Law of Probability - Simple Case

If two events $A$ and $B$ are mutually exclusive then

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

Key Point 6

The Addition Law of Probability - General Case

If two events are $A$ and $B$ then

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

A \cap B = \emptyset

, i.e.

A

and

B

are mutually exclusive, then

P (A \cap B) = P (\emptyset) = 0

, and this general expression reduces to the simpler case.

This rule can be extended to three or more events, for example:

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

Example 10

Consider a pack of 52 playing cards. A card is selected at random. What is the probability that the card is either a diamond or a ten?

Solution

If $A$ is the event { a diamond is selected } and $B$ is the event { a ten is selected } then obviously $P (A) = \frac{13}{52}$ and $P (B) = \frac{4}{52}$ . The intersection event $A \cap B$ consists of only one member - the ten of diamonds - which gets counted twice hence $P (A \cap B) = \frac{1}{52}$ .

Therefore $P (A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}$ .

Task!

A bag contains 20 balls, 3 are coloured red, 6 are coloured green, 4 are coloured blue, 2 are coloured white and 5 are coloured yellow. One ball is selected at random. Find the probabilities of the following events.

the ball is either red or green
the ball is not blue
the ball is either red or white or blue. (Hint: consider the complementary event.)

Note that a ball has only one colour, designated by the letters $R, G, B, W, Y$ .

$P (R \cup G) = P (R) + P (G) = \frac{3}{20} + \frac{6}{20} = \frac{9}{20} .$
$P (B^{'}) = 1 - P (B) = 1 - \frac{4}{20} = \frac{16}{20} = \frac{4}{5} .$
The complementary event is $G \cup Y$ , $P (G \cup Y) = \frac{6}{20} + \frac{5}{20} = \frac{11}{20}$ .
Hence $P (R \cup W \cup B) = 1 - \frac{11}{20} = \frac{9}{20}$

In the last Task part (c) we could alternatively have used an obvious extension of the law of addition for mutually exclusive events:

$P (R \cup W \cup B) = P (R) + P (W) + P (B) = \frac{3}{20} + \frac{2}{20} + \frac{4}{20} = \frac{9}{20} .$

Task!

The diagram shows a simplified circuit in which two independent components $a$ and $b$ are connected in parallel.

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The circuit functions if either or both of the components are operational. It is known that if $A$ is the event ‘component $a$ is operating’ and $B$ is the event ‘component $b$ is operating’ then $P (A) = 0.99$ , $P (B) = 0.98$ and $P (A \cap B) = 0.9702$ . Find the probability that the circuit is functioning.

The probability that the circuit is functioning is $P (A \cup B)$ . In words: either $a$ or $b$ or both must be functioning if the circuit is to function. Using the keypoint:

\begin{array}{rcl} P (A \cup B) & = & P (A) + P (B) - P (A \cap B) \\ = & 0.99 + 0.98 - 0.9702 = 0.9998 \end{array}

Not surprisingly the probability that the circuit functions is greater than the probability that either of the individual components functions.

Exercises

The following people are in a room: 5 men aged 21 and over, 4 men under 21, 6 women aged 21 and over, and 3 women under 21. One person is chosen at random. The following events are defined: A = { the person is aged 21 and over } ; B = { the person is under 21 } ; C = { the person is male } ; D = { the person is female } . Evalute the following:
1. $P (B \cup D)$
2. $P (A^{'} \cap C^{'})$
  
  Express the meaning of these events in words.
A card is drawn at random from a deck of 52 playing cards. What is the probability that it is an ace or a face card (i.e. K, Q, J)?
In a single throw of two dice, what is the probability that neither a double nor a sum of 9 will appear?

(a) $P (B \cup D) = P (B) + P (D) - P (B \cap D)$
$P (B) = \frac{7}{18}, P (D) = \frac{9}{18}, P (B \cap D) = \frac{3}{18}$

$∴ P (B \cup D) = \frac{7}{18} + \frac{9}{18} - \frac{3}{18} = \frac{13}{18}$

(b) $P (A^{'} \cap C^{'}) A^{'} = {people under 21} C^{'} = {people who are female}$

$∴ P (A^{'} \cap C^{'}) = \frac{3}{18} = \frac{1}{6}$
$F = {face card} A = {card is ace} P (F) = \frac{12}{52}, P (A) = \frac{4}{52}$
$∴ P (F \cup A) = P (F) + P (A) - P (F \cap A) = \frac{12}{52} + \frac{4}{52} - 0 = \frac{16}{52}$
$D = {double is thrown} N = {sum is 9}$
$P (D) = \frac{6}{36}$ (36 possible outcomes in an experiment in which all the outcomes are equally probable).

$P (N) = P {(6 \cap 3) \cup (5 \cap 4) \cup (4 \cap 5) \cup (3 \cap 6)} = \frac{4}{36}$

$P (D \cup N) = P (D) + P (N) - P (D \cap N) = \frac{6}{36} + \frac{4}{36} - 0 = \frac{10}{36}$

$P ({(D \cup N)}^{'}) = 1 - P (D \cup N) = 1 - \frac{10}{36} = \frac{26}{36}$

1 The addition law

1.1 Types of events

1.2 Mutually exclusive events

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