Exercise 4 Page 305

We are asked how to find the probabilities of disjoint and overlapping events. Let $A$ and $B$ be two events. We can use a Venn diagram to illustrate the probabilities of disjoint and overlapping events. In both cases, we find the individual probabilities, $P(A)$ or $P(B),$ by dividing the favorable outcomes by the total number of outcomes.

Disjoint Events

We can see that the probabilities of disjoint events do not share any region in the diagram. This means that there is no favorable outcome that satisfies both $A$ and $B.$ Therefore, is impossible for $A$ and $B$ to occur at the same time. Let's write this as a probability. Similarly, since disjoint events do not share any outcomes, to find the probability that $A$ or $B$ occurs, we can add their probabilities.

Overlapping Events

We can see that overlapping events share a region in the diagram. This is because there is at least one possible outcome that is favorable for both $A$ and $B.$ Therefore, the probability that $A$ and $B$ occur is always greater than $0.$ Looking at the diagram, we can see that if we only added the probabilities of $A$ and $B,$ we would add the probability $P(A\text{ and }B)$ twice. Because of this, we need to subtract this probability from the sum to find the probability that $A$ or $B$ occurs.

Example

Let's consider some possible events that could occur when using a standard deck of cards.

Disjoint Events Example

First, let's say that we want to know the probability that we will randomly draw a $2$ or a $10.$ Since there are no favorable outcomes that would satisfy both events, these are disjoint events. Let's draw these events with a Venn diagram!

Since there are $52$ cards in a deck and $4$ types of cards, we will divide $4$ by $52$ to find the probability of each individual event. We know that there are no favorable outcomes for both events. Therefore, the probability of drawing a card with a $2$ and a $10$ is $0.$ Now we will find the probability that we draw either a $10$ or a $2.$ Since the events are disjoint, we will find this probability by adding the individual probabilities. We can confirm this probability seeing that $8$ of the $52$ cards are favorable outcomes.

Overlapping Events Example

Now consider the events of drawing a $10$ and a spade.

The probability of drawing a $10$ is the same as in the previous example, $\frac{4}{52}.$ Since there are $13$ of each type of card, we find the probability of drawing a spade by dividing $13$ by $52.$ We can see that there is a favorable outcome that satisfies both events: drawing the $10$ of spades. Let's find the probability of drawing a card that is a $10$ and a spade. Now let's find the probability of drawing a $10$ or drawing a spade. Looking at the diagram, we can see that there are $16$ cards that are favorable outcomes for this event, confirming this probability.