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,

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,
$$\textmd{Since }P(X \cap Y) = P(X) \times P(Y)\textmd{, then events X and Y are independent}$$