The Covid-19 pandemic caused a lot of businesses to crumble and people to lose their jobs. This led to people building businesses that could still thrive during the pandemic. We can say that these businesses are independent of the pandemic.
This is what independent events are. The business is an event and Covid-19 is another and they have no effect on each other.
In this article, we will see the definition of independent events, formulas related to independent events and examples of their application. We will also see how we can visually represent this type of events in the form of what is known as Venn diagrams.
Independent events definition
An Independent event is when the occurrence of one event does not influence the probability of another event happening.
You can have two separate events that have nothing to do with each other. Whether one occurs or not will not affect the behaviour of the other. That's why they are called independent events.
When you toss a coin you get either heads or tails. Perhaps you've tossed the coin three times and it landed on heads those three times. You might think there's a chance for it to land on tails when you toss it the fourth time, but that's not true.
The fact that it has been landing on heads doesn't mean that you might get lucky and get a tail next time. Getting heads and getting a tail when a coin is tossed are two independent events.
Suppose you're buying a car and your sister hopes to get into a university. In that case, these two events are also independent, because your buying a car will not affect your sister's chances of getting into a university.
Other examples of independent events are:
Winning the lottery and getting a new job;
Going to college and getting married;
Winning a race and getting an engineering degree.
There are times when it may be challenging to know if two events are independent of each other. You should take note of the following when trying to know if two (or more) events are independent or not:
Independent events probability formula
To find the probability of an event happening, the formula to use is:
\[\text{Probability of an event happening} = \frac{\text{Number of ways the event can happen}}{\text{Number of possible outcomes}}\]
Here, we are talking about independent events probabilities and you may want to find the probability of two independent events happening at the same time. This is the probability of their intersection. To do this, you should multiply the probability of one event happening by the probability of the other. The formula to use for this is below.
\[P(A \space and \space B) = P(A \cap B) = P(A) \cdot P(B)\]
where P is probability
\(P (A \cap B)\) is the probability of the intersection of A and B
P(A) is the probability of A P(B) is the probability of B
Consider independent events A and B. P(A) is 0.7 and P(B) is 0.5, then:
\(P(A \cap B) = 0.7 \cdot 0.5 = 0.35\)
This formula can also be used to find out if two events are indeed independent of each other. If the probability of the intersection is equal to the product of the probability of the individual events, then they are independent events otherwise they are not.
We will look at more examples later.
Independent events represented in Venn diagrams
A Venn diagram is for visualisation purposes. Recall the formula for finding the probability of two independent events happening at the same time.
\[P(A \cap B) = P(A) \cdot P(B)\]
The intersection of A and B can be shown in a Venn diagram. Let's see how.
A Venn diagram - StudySmarter Original
The Venn diagram above shows two circles representing two independent events A and B that intersect. S represents the entire space, known as sample space. The Venn diagram gives a good representation of the events and it may help you understand the formulas and calculations better.
The sample space represents the possible outcomes of the event.
When drawing a Venn diagram, you may need to find the probability of the entire space. The formula below will help you do that.
\[S = 1 - (P(A) + P(A \cap B) + P(B))\]
Independent events probability examples and calculations
Let's put the formulas we've talked about to use in the examples below.
Consider two independent events A and B that involve rolling a die. Event A is rolling an even number and event B is rolling a multiple of 2. What is the probability of both events happening at the same time?
Solution
We have two events A and B.
Event A - rolling an even number
Event B - rolling a multiple of 2
Both events are independent. A die has six sides and the possible numbers to appear are 1, 2, 3, 4, 5, and 6. We are asked to find the probability of both events happening at the same time which is the intersection of both.
The formula to use is:
\(P (A \cap B) = P (A) \cdot P(B)\)
From the formula, we can see that to calculate the intersection, you need to know the probability of each event happening.
\[\text{Probability of an event happening} = \frac{\text{Number of ways the event can happen}}{\text{Number of possible outcomes}}\]
Therefore
\(P(A) = \frac{3}{6} = \frac{1}{2}\)
\(P(B) = \frac{3}{6} = \frac{1}{2}\)
We will now substitute the formula
\(P (A \cap B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\)
So the probability of both events happening is \(\frac{1}{4}\).
Let's take another example.
\(P(A) = 0.80\) and \(P(B) = 0.30\) and A and B are independent events. What is \(P(A \cap B)\)?
Solution
We are asked to find \(P(A \cap B)\) when \(P(A) = 0.80\) and \(P(B) = 0.30\). All we have to do is substitute into the formula below.
\(P (A \cap B) = P(A) \cdot P(B) = 0.80 \cdot 0.30\)
\(P(A \cap B) = P(A) \cdot P(B) = 0.80 \cdot 0.30\)
Therefore, \(P(A \cap B) = 0.24\)
In a classroom, 65% of the students like mathematics. If two students are chosen at random, what is the probability that both of them like mathematics and what is the probability that the first student likes mathematics and the second doesn't?
Solution
We have two questions here. The first is to find the probability of both students liking mathematics and the other is to find the probability of one liking mathematics and the other not liking it.
One student liking mathematics does not have affect on whether the second student likes mathematics too. So they are independent events. The probability of both of them liking mathematics is the probability of the intersection of the events.
If we call the events A and B, we can calculate using the formula below.
\(P(A \cap B) = P(A) \cdot P(B) = \frac{65}{100} \cdot \frac{65}{100}\)
Notice we divided by 100. This is because we are dealing with percentages.
Now, to find the probability of the first student liking mathematics and the second not liking it. These two are separate independent events and to find what we are looking for, we have to find the intersection of both events.
The probability of the first student liking mathematics is
\(P(A) = 65\% = 0.65\)
The probability of the second student not liking mathematics is
\(P(B) = 1- 0.65 = 0.35\)
We will now get our final answer by substituting the equation above.
\(P(A \cap B) = P(A) \cdot P(B) = 0.65 \cdot 0.35\)
Let's see a fourth example.
C and D are events where \(P(C) = 0.50, \space P(D) = 0.90\). If \(P(C \cap D) = 0.60\), are C and D independent events?
Solution
We want to know if events C and D are independent. To know this, we will use the formula below.
\(P(C \cap D) = P(C) \cdot P(D)\)
We are given
\(P(C) = 0.50 \quad P(D) = 0.90 \quad P(C \cap D) = 0.60\)
If we substitute in the formula and we get the intersection to be something different from what the question suggests, then the events are not independent otherwise, they are independent.
Let's substitute.
\(P(C \cap D) = 0.50 \cdot 0.90 \quad P(C \cap D) = 0.45\)
We got 0.45 and the question says the intersection should be 0.60. This means the events are not independent.
Next, the fifth example.
A and B are independent events where \(P(A) = 0.2\) and \(P(B) = 0.5\). Draw a Venn diagram showing the probabilities for the event.
Solution
The Venn diagram needs some information to be put in it. Some of them have been given and we have to calculate for others.
\(P(A) = 0.2 \quad P(B) = 0.5 \quad P(A \cap B) = ? \quad P(S) = ? \space \text{(probability of the entire space)}\)
Now let's find the missing information.
\(P(A \cap B) = P(A) \cdot P(B) = 0.2 \cdot 0.5 = 0.1\)
\(P(S) = 1 - (P(A) + P(A \cap B) + P(B)) = 1-(0.2 + 0.1 +0.5) = 1-0.8 = 0.2\)
Now, let's draw the Venn diagram and put in the information.
From the Venn diagram below, find
- \(P(C \cap D)\)
- \(P(C \cup D)\)
- \(P(C \cup D')\)
Solution
a. \(P(C \cap D)\)
\(P(C \cap D) = P(C) \cdot P(D)\)
From the Venn diagram,
\(P(C) = 0.2 \quad P(D) = 0.6\)
So we will now substitute the formula.
\(P(C \cap D) = P(C) \cdot P(D) = 0.2 \cdot 0.6 = 0.12\)
b. \(P(C \cup D)\)
Here, we are to find the union of both events. This will be the summation of the probability of C, D and the intersect.
\(P(C \cup D) = P(C) + P(D) +P(C \cup D) = 0.2 + 0.6 + 0.12\)
c. \(P(C \cup D')\)
\(C \cup D'\) means everything in C that is not in D. If we look at the Venn diagram, we will see that this comprises 0.2, \(C \cap D\) and 0.8.
So we have:
\(P(C \cup D') = P(C) + P(C \cap D) + S = 0.2 +0.12 + 0.08 = 0.4\)
Independent Probabilities - Key takeaways
- Independent event probability is when the occurrence of one event does not influence on the probability of another event happening.
- The formula for calculating the probability of two events happening at the same time is:
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- The formula for calculating the probability of two events happening can also be used to find out if two events are indeed independent of each other. If the probability of the intersection is equal to the product of the probability of the individual events, then they are independent events otherwise they are not.
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