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Probability Of Multiple Events – Conditions, Formulas, and Examples
The probability of multiple events is an interesting topic discussed in mathematics and statistics. There are instances where we’re observing multiple events and want particular results – when this happens, knowing how to calculate the probability of multiple events comes in handy.
The probability of multiple events helps us measure our chances of getting the desired outcomes when two or more vents are occurring. The measured probability will heavily depend on whether the given events are independent or dependent.
Seeing that this is a more complex topic than the earlier topics of probability, make sure to refresher your knowledge on the following:
Understand how we calculate probabilities of a single event.
Review what complementary probabilities are.
Let’s begin by understanding when we apply the particular probability we’re discussing – and we can do so by studying the spinner shown in the next section.
What are multiple events in probability?
The probability of multiple events occurs when we’re trying to calculate the probability of observing two or more events. These include experiments where we’re observing different behaviors simultaneously, drawing cards with multiple conditions, or predicting the outcome of a multi-colored spinner.
Speaking of spinners, why don’t we observe the image shown above? From this, we can see that the spinner is divided into seven regions and distinguished by either the region’s colors or labels.
Here are examples of multiple events we can check from the spinners:
Finding the probability of spinning a violet or an $a$.
Finding the probability of spinning a blue or a $b$.
These two conditions will require us to calculate the probability of two events occurring at the same time.
Multiple events probability definition
Let’s dive right into the definition of multiple event probabilities and when they occur. The probability of multiple events measures the likelihood that two or more events occur at the same time. We sometimes lookout for the probability of when one or two outcomes happen and whether these outcomes overlap each other.
The probability will depend on an important factor: whether the multiple events are independent or not and whether they are mutually exclusive.
Dependent events (also known as conditional events) are events where a given event’s outcomes are affected by the remaining events’ outcomes.
Independent events are events where one event’s outcomes are not affected by the rest of the events’ outcomes.
Here are some examples of events that are dependent and independent of each other.
Dependent Events | Independent Events |
Drawing two balls consecutively from the same bag. | Finding one ball each from two bags. |
Picking two cards without replacement. | Picking a card and rolling a die. |
Buying more lottery tickets to win the lottery. | Winning the lottery and seeing your favorite show on a streaming platform. |
Events can also be mutually exclusive– these are events where they can never happen simultaneously. Some examples of mutually exclusive are the chances of turning left or right at the same time. Ace and king cards from a deck are also mutually exclusive.
Knowing how to distinguish these two events will be extremely helpful when we learn how to evaluate the probabilities of two or more events that occur together.
How to find the probability of multiple events?
We’ll be using different approaches when finding the probability of multiple events occurring together depending on whether these events are dependent, independent, or mutually exclusive.
Finding the Probability of Independent Events
\begin{aligned}P(A \text{ and } B) &=P(A) \times P(B)\\P(A \text{ and } B \text{ and } C\text{ and }…) &=P(A) \times P(B) \times P(C) \times … \end{aligned}
When we’re working with independent events, we can calculate the probability occurring together by multiplying the respective probabilities of the events occurring individually.
Let’s say we have the following objects handy:
A bag that contains $6$ red and $8$ blue chips.
A coin is in your purse.
A deck of cards is on your office table.
How do we find the probability that we get a red chip and toss the coin and get tails, and draw a card with a heart suit?
These three events are independent of each other, and we can find the probability of these events occurring together by first finding the probability that they occur independently.
As a refresher, we can find their independent probabilities by dividing the number of outcomes by the total number of possible outcomes.
Event | Symbol | Probability |
Getting a red chip | $P(r)$ | $P(r) = \dfrac{6}{14} = \dfrac{5}{7}$ |
Tossing the coin and get a tails | $P(t)$ | $P(t) = \dfrac{1}{2}$ |
Drawing a hearts | $P(h)$ | $P(h) = \dfrac{13}{52} = \dfrac{1}{4}$ |
\begin{aligned}P(r \text{ and }t \text{ and }h)&= P(r) \cdot P(t)\cdot P(h)\\&= \dfrac{5}{7}\cdot \dfrac{1}{2} \cdot \dfrac{1}{4}\\&= \dfrac{5}{56} \end{aligned} |
Finding the Probability of Dependent Events
\begin{aligned}P(A \text{ and } B) &=P(A) \times P(B \text{ given } A)\\&= P(A)\times P(B|A)\\P(A \text{ and } B \text{ and } C) &=P(A) \times P(B \text{ given } A)\times P(C \text{ given } A\text{ and }B)\\&=P(A) \times P(B|A)\times P(C|A \text{ and } B) \end{aligned}
We can calculate for the probability of dependent events occurring together as shown above. Need a refresher on what $P(A|B)$ represents? It simply means the probability of $A$, once $B$ has happened. You’ll know more about conditional probability and be able to try out more complex examples here.
Let’s say we want to find out the probability of getting three jacks consecutively if we don’t return the drawn card each draw. We can keep in mind that three events are occurring in this situation:
The probability of getting a jack on the first draw – we still have $52$ cards here.
The probability of getting a second jack on the second draw (we now have $3$ jacks and $51$ cards).
The third event is getting a third jack for the third row – $2$ jacks left and $50$ cards on the deck.
We can label these three events as $P(J_1)$, $P(J_2)$, and $P(J_3)$. Let’s work on the important components to calculate the probability of these three dependent events happening together.
Event | Symbol | Probability |
Drawing a jack the first time | $P(J_1)$ | $\dfrac{4}{52}= \dfrac{1}{13}$ |
Drawing a jack the second time | $P(J_2|J_1)$ | $\dfrac{4 -1}{52 -1} = \dfrac{1}{17}$ |
Drawing a jack the third time | $P(J_3|J_1 \text{ and } J_2)$ | $\dfrac{3-1}{51 -1} = \dfrac{1}{25}$ |
\begin{aligned}P(J_1) \times P(J_2 \text{ given } J_1)\times P(J_3 \text{ given } J_2\text{ and }J_1)&=P(J_1) \times P(J_2|J_1)\times P(J_3|J_1 \text{ and } J_2)\\&=\dfrac{4}{52}\cdot\dfrac{3}{51}\cdot\dfrac{2}{50}\\&= \dfrac{1}{13}\cdot \dfrac{1}{17}\cdot \dfrac{1}{25}\\&= \dfrac{1}{5525} \end{aligned} |
Finding the Probability of Mutually Exclusive or Inclusive Events
We may also need to explore if the given events are mutually inclusive or exclusive to help us calculate the probability of multiple events where the outcome we’re looking for do not require all outcomes to occur altogether.
Here’s a table that summarizes the formula for mutually exclusive or inclusive events:
Type of Event | Formula for the Probability |
Mutually Inclusive | $P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$ |
Mutually Exclusive | $P(A \text{ or } B) = P(A) + P(B)$ |
Keep in mind that we’re now using “or” because we’re looking for the probabilities of events that occur individually or occur together.
These are all the concepts and formulas you’ll need to understand and solve problems that involve multiple events’ probability. We can go ahead and try out these examples shown below!
Example 1
A canvas bag contains $6$ pink cubes, $8$ green cubes, and $10$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is pink, the second cube is purple, and the third is another pink cube?
Solution
Keep in mind that the cubes are returned each time we draw another. Since the next draw’s probability is not affected by the first draw results, the three events are independent of each other.
When this happens, we multiply the individual probabilities to find the probability of having the outcome that we want.
Event | Symbol | Probability |
Drawing a pink cube in the first draw | $P(C)$ | $P(C_1) = \dfrac{6}{24}= \dfrac{1}{4}$ |
Drawing a purple cube in the second draw | $P(C_2)$ | $P(C_2) = \dfrac{10}{24}= \dfrac{5}{12}$ |
Drawing another pink cube in the third draw | $P(C_3)$ | $P(C_3) = \dfrac{6}{24}= \dfrac{1}{4}$ |
\begin{aligned}P(C_1 \text{ and }C_2\text{ and }C_3)&= P(C_1) \cdot P(C_2)\cdot P(C_3)\\&= \dfrac{1}{4}\cdot \dfrac{5}{12} \cdot \dfrac{1}{4}\\&= \dfrac{5}{192} \end{aligned} |
This means that the probability of drawing a pink cube then a purple cube then another pink cube is equal to $\dfrac{5}{192}$.
Example 2
A book club of $40$ enthusiastic readers, $10$ prefers nonfiction books, and $30$ prefers fiction. Three book club members will be randomly selected to serve as the next book club meeting’s three hosts. What is the probability that all three members will prefer nonfiction?
Solution
When the first member is selected as the first host, we can no longer include them in the next random selection. This shows that the three outcomes are dependent on each other.
For the first selection, we have $40$ members and $30$ nonfiction readers.
For the second selection, we now have $40 -1 = 39$ members and $30- 1= 29$ nonfiction readers.
Hence, for the third, we have $38$ members and $28$ nonfiction readers.
Event | Symbol | Probability |
Randomly selecting a nonfiction reader | $P(N_1)$ | $\dfrac{30}{40}= \dfrac{3}{4}$ |
Selecting another nonfiction reader | $P(N_2|N_1)$ | $\dfrac{29}{39}$ |
Selecting a nonfiction reader the third time | $P(N_3|N_1 \text{ and } N_2)$ | $\dfrac{28}{38} = \dfrac{14}{19}$ |
\begin{aligned}P(N_1) \times P(N_2 \text{ given } N_1)\times P(N_3 \text{ given }N_2\text{ and }N_1)&=P(N_1) \times P(N_2|N_1)\times P(N_3|N_1 \text{ and } N_2)\\&=\dfrac{30}{40}\cdot\dfrac{29}{39}\cdot\dfrac{28}{38}\\&= \dfrac{3}{4}\cdot \dfrac{29}{39}\cdot \dfrac{14}{19}\\&= \dfrac{203}{494} \end{aligned} |
Hence, the probability of selecting three nonfiction readers is equal to $\dfrac{203}{494}\approx 0.411$.
Example 3
Let’s go back to the spinner that was introduced to us in the first section, and we can actually determine the probabilities of the following:
a. Spinning a violet or an $a$.
b. Spinning a blue or a red.
Solution
Let’s take note of the colors and labels found in each spinner.
Color $\rightarrow$ Label $\downarrow$ | Violet | Green | Red | Blue | Total |
$a$ | $1$ | $1$ | $0$ | $1$ | $3$ |
$b$ | $2$ | $0$ | $0$ | $0$ | $2$ |
$c$ | $0$ | $0$ | $1$ | $1$ | $2$ |
Total | $3$ | $1$ | $1$ | $2$ | $7$ |
Take note of the keyword “or” – this means that we account for the probability that either outcome occurs. For problems like this, it’s important to note whether the conditions are mutually exclusive or inclusive.
For the first condition, we want the spinner to land on either a violet region or a region labeled $a$, or both.
There are $3$ violet regions and $3$ regions labelled $a$.
There is a $1$ region where it’s both violet and labeled $a$.
This shows that the incident is mutually inclusive. Hence, we use $P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$
\begin{aligned}P(V \text{ or } a) &= P(V) + P(a) – P(V \text{ and } a)\\&=\dfrac{3}{7} + \dfrac{3}{7} – \dfrac{1}{7}\\&= \dfrac{5}{7}\end{aligned}
a. This means that the probability is equal to $\dfrac{5}{7}$.
It’s impossible to land on a red region and a blue one all at the same time. This means that these two events are mutually exclusive. For these types of events, we add their individual probabilities.
b. This means that the probability is equal to $\dfrac{1}{7} + \dfrac{2}{7} = \dfrac{3}{7}$.