5.3.2 Exclusive and Independent Events

In this topic we will learn how to:

understand the meaning of exclusive and independent events, including the determination of whether events $A$ and $B$ are independent by comparing the values of $P(A \cap B)$ and $P(A) \times P(B)$

$\textbf{\textcolor{gray}{Mutually exclusive Events}}$ Exclusive events are ones that cannot occur at the same time. For example, if a fair die is tossed, it is not possible to get both a $3$ and a $6$ at the same time. Therefore, event A ‘tossing a $3$ ‘ cannot occur at the same time as event B ‘tossing a $6$ ‘. Events A and B are said to be mutually exclusive since they cannot occur at the same time. If events A and B are mutually exclusive then we can say,
$P(A \cap B) = 0$ Since they do not intersect and do not occur at the same time. $\textbf{\textcolor{gray}{Independent Events}}$ Independent events are events whose occurrence is not dependent on another event. If events are independent they can occur at the same time. An example is, if using a fair die, you roll a $2$ , then on your second roll you get a $5$ . The occurence of the two events is independent of the other. Rolling a $2$ on your first roll does not increase or decrease your chances of rolling a $5$ on the second roll. If events A and B are independent, then,
$P(A \cap B) = P(A) \times P(B)$ Let’s look at some past paper questions.

1. There are $400$ students at a school in a certain country. Each student was asked whether they preferred swimming, cycling or running and the results are given in the following table. (9709/52/F/M/21 number 7)

(a) A student is chosen at random. Determine whether the events ‘the student is male’ and ‘the student prefers swimming’ are independent, justifying your answers.
$\textbf{Let event A be 'the student is male'}$ $\textbf{Let event B be 'the student prefers swimming'}$ If the events A and B are independent then,
$P(A \cap B) = P(A) \times P(B)$ Use the table to find the probability of event A happening,
$P(A) = \frac{31 + 57 + 92}{400}$ $P(A) = \textcolor{#2192ff}{\frac{9}{20}}$ Use the table to find the probability of event B happening,
$P(B) = \frac{104 + 31}{400}$ $P(B) = \textcolor{#0f0}{\frac{27}{80}}$ Use the table to find the probability of events A and B intersecting,
$P(A \cap B) = \textcolor{red}{\frac{31}{400}}$ Now let’s go back to the idea that if events A and B are independent then,
$P(A \cap B) = P(A) \times P(B)$ $\textcolor{red}{\frac{31}{400}} \neq \textcolor{#2192ff}{\frac{9}{20}} \times \textcolor{#0f0}{\frac{27}{80}}$ $\frac{31}{400} \neq \frac{243}{1\ 600}$ Therefore, the final answer is,
$\textmd{Since }P(A \cap B) \neq P(A) \times P(B)\textmd{ then events A and B}\\ \textmd{ are not independent of each other}$ 2. A fair six-sided die is thrown twice and the scores are noted. Event X is defined as ‘The total of the two scores is $4$ ‘. Event Y is defined as ‘The first score is $2$ or $5$ ‘. Are events X and Y independent? Justify your answer. (9709/61/M/J/19 number 3)

To be able to solve this question we have to draw a possibility space diagram. Label the $y$ -axis with, Throw 1. Label the $x$ -axis with, Throw 2. Since we are using a fair die, label both axes with ticks from $1 - 6$ . For each possible outcome plot a dot. Identify and mark events X and Y on the diagram. Define a key to explain how you labeled each event on the diagram. Below is an example of a possibility space diagram relevant to this question,

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From the possibility space diagram, we can tell that,
$P(X) = \frac{3}{36}$ $P(Y) = \frac{12}{36}$ $P(X \cap Y) = \frac{1}{36}$ If events X and Y are independent, then,
$P(X \cap Y) = P(X) \times P(Y)$ $\frac{1}{36} = \frac{3}{36} \times \frac{12}{36}$ $\frac{1}{36} = \frac{1}{36}$ Therefore, the final answer is,
$\textmd{Since }P(X \cap Y) = P(X) \times P(Y)\textmd{, then events X and Y are independent}$