Dependent events
A juggler enthralled a circus audience with his skills.
In his hands, he has a red club, two green clubs, and three blue clubs. His juggling technique is such that he does not use the same club twice.
During an act, he picks one club up and throws it into the air, and then he picks a second one up and throws it after catching the first one.
What is the probability that the first juggling club is blue and the second juggling club is green?
Let us understand this concept better.
In this chapter, we will learn about dependent events and how to find the probability of dependent events. As independent events are a part of probability, we also learn the difference between independent and dependent events.
Lesson Plan
What Are Dependent Events?
Two events are said to be dependent if the outcome of one event affects the outcome of the other.
In probability, dependent events are usually real-life events and rely on another event to occur. For example, Sam scored well in his math test because he studied for it; the gym class had a football session because Adam got a football from home. If you look at these examples, then you will notice that one event is dependent on the other from happening.
Mathematically we represent the dependent events in probability
As we have learned about the probability that:
\[\text{Probability} = \dfrac{\text{Favourable Outcome}}{\text{Possible Outcome}} \]
NOTE:
\(\text{P(A)}\) represents the probability of happening an event \(\text A \)
\(\text{P(B)}\) represents the probability of happening an event \(\text B \)
The Probability of Dependent Events
If A and B are dependent events, then the probability of A and B occurring is written as:
Given, Probability of event A is \(\text{P(A)}\)
Probability of event B is\[ \text{ P(B after A)}\]
\[ \text{P(B and A)} = \text{P(A)}\times\text{P(B after A)}\].
\( \text{P(B after A)}\) can also be written as \(\text P(B | A)\).
\(\text P(B | A)\) means that event A has already happened.
Now, what is the chance of happening of event B?
\(\text P(B | A)\) is also called the "Conditional Probability" of B given A.
then \( \text{P(B and A)} = \text{P(A)} \times \text P(B | A)\).
Difference Between Independent and Dependent Events
There are two types of events in probability which are often classified as dependent or independent events.
The difference is given below in the table.
| Dependent Events | Independent Events |
|---|---|
| 1. The occurrence of one event affecting the probability of another event. | 1. The occurrence of one event not affecting the probability of another event. |
|
2.Examples include a power cut in case you don't pay your bill on time, winning the lottery after buying 10 lottery tickets (the more the tickets bought, the greater the chance of winning) |
Examples include riding a bike and watching your favorite movie on a laptop
|
|
3. Formula can be written as: |
3. Formula can be written as: |
How Does One Find The Probability of Dependent Events?
To find the probability of dependent events, one uses the formula for conditional probability given below:
If the probability of events A and B is P(A) and P(B) respectively then the conditional probability of event B such that event A has already occurred is P(B/A).
The formula to calculate conditional probability.
\[ P\left( \dfrac BA \right)=\dfrac {P(A \cap B)}{P(A)} \text {or} \dfrac {P(B \cap A)}{P(A)}\]
Given, P(A) must be greater than 0.
P(A) less than 0 means A is an impossible event. In P(A \(\cap\) B) the intersection denotes a compound probability of an event.
Let's Find the Probability of Dependent Events Through an Example in Detail
Consider a box with 10 toys. Of these, seven are multicolored, and three are blue.
Based on this information, there is a 7 out of 10 chance of pulling a multicolored toy from the box.
Similarly, there is a 3 out of 10 chance of pulling a blue toy out of the box.
However, if we randomly select two toys from the box, then what is the probability that we pull out a multicolored toy and then a blue toy without putting it back in the box?
A common error while solving such problems is using the formula and then multiplying the probability of each toy together.
As the toys are being taken out without putting back in the box, it means that the probability will change after every draw.
This situation shows that an event is dependent on another event.
Let's go back to the original situation. If we pull a multicolored toy out of the box that contains 10 toys, then the chances of the toy being multicolored are 7 out of 10.
In the second event, the probability of pulling out a blue toy is, however, not 3 out of 10, as one multicolored toy is not put back in the box.
The probability is now 3 out of 9 toys.
To finish this problem, all we have to do is multiplied this value together \( \dfrac 7{10} \times \dfrac 39 = \dfrac{21}{90} = \dfrac 7{30} = 0.233 \).
Few Steps to Check Whether the Probability Belongs to Dependent or Independent Events
Step 1: Is it possible for the events to happen in any order? If yes, go to step 2, if no, go to step 3.
Step 2: Does one event affect the outcome of the other event? If yes, go to step 4, if no, go to step 3.
Step 3: The event is independent. Simply put the formula of independent event and get the answer.
Step 4: The event is dependent. Simply put the formula of dependent event and get the answer.
That’s how to find out if an event is Dependent or Independent!
- The probabilities of an event that do not affect one another with replacement are independent.
- The probabilities of an event that do affect one another without replacement are dependent.
Solved Examples
| Example 1 |
Daniel has drawn a card from a well-shuffled deck. Find the probability of the card either being red or a king card.
Solution
Let us define the event E as the card drawn is either red or a king card.
How many outcomes are there which are favorable to E?
There are 26 red cards, and 4 cards which are kings. However, 2 of the red cards are themselves, kings.
If we add 26 and 4, we will be counting these two cards twice.
Thus, the correct number of outcomes which are favorable to E is-
\[ 26 + 4 − 2 = 28\]
The probability of E occurring will be-
\( \text{ P(E) }= \dfrac {28}{52} = \dfrac{7}{13} \)
| \(\therefore\) The probability of the card either being red or a king card is \(\dfrac{7}{13}\) |
| Example 2 |
A juggler has seven red, five green, and four blue balls. During his stunt, he accidentally drops a ball and doesn't pick it up. As he continues, another ball falls down. What is the probability that the first ball that was dropped is blue, and the second ball is green?
Solution
As we know that the first ball is not replaced by the juggler. So after dropping the first ball, he is left with 15 balls.
The probability that the first ball is blue or P(blue ball) = \(\dfrac{4}{16}\)
The probability that the second ball is green or P(green ball) = \(\dfrac{5}{15 }\)
The probability that the first ball is blue and the second ball is green:
\( \text {P(blue than green) = P(blue)}\times\text {P(green) } \)
\(=\dfrac{4}{16}\times \dfrac{5}{15 } =\dfrac{1}{12 }\)
| \(\therefore\) The probability that the first ball is blue and the second ball is green is \(\dfrac{1}{12}\) |
| Example 3 |
Mrs. Andrews has to select two students from 35 girls and 15 boys to be part of a club. What is the probability that both students are girls?
Solution
The total number of students \(= 35 + 15 = 50\)
Probability of choosing the first girl, P(girl 1) = \( \dfrac { 35}{50}\)
Probability of choosing the second girl, P(girl 2) = \( \dfrac { 34}{49}\)
Now,
The probability that both students have chosen are girls
\(\text {P(first girl and second girl)} \)
\(= \text{ P(first girl)} \times \text{P(second girl | first girl)} \)
\(= \dfrac { 35}{50} \times \dfrac { 34}{49} \)
\(= \dfrac{1190}{1666}\)
\(= \dfrac{85}{119}\ \)
| \(\therefore \)The probability that both chosen students are girls is \(\dfrac{85}{119}\). |
| Example 4 |
Joseph and David are playing a game of cards. Joseph drew a card at random without replacement. He asks David to help him determine the probability that the first card drawn was a queen and the second is a king.
Solution
As we understand that this probability is having a dependent event condition.
P (drawing a queen in the first condition) =\( \dfrac { 4}{52}\)
P (drawing a king in the second condition after a queen) = \( \dfrac { 4}{51}\)
P (drawing a queen followed by a king) = \(\dfrac{4}{52} \times\dfrac{4}{51}=\dfrac{16}{2652}=\dfrac{4}{663}\)
| \(\therefore \)The probability of drawing a queen followed by a king is \(\dfrac{4}{663}\). |
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80% of your friends like burgers, and 50% like burgers and pizza.
What percent of those like burgers but not pizza?
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What is the probability of drawing 4 kings from a deck of cards?
Interactive Questions
Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
Let's Summarize
This mini-lesson targeted the fascinating concept of Dependent Events. The math journey around the octahedron starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
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Frequently Asked Questions (FAQs)
1. How does one determine if events are independent or dependent?
In probability, if one event affects the outcome of the other event, is called a dependent event but if one event does not affect the outcome of the other event that event is called independent.
2. What is an event in probability?
In probability, an event is an outcome of an experiment or an event is said to be a set of outcomes of an experiment to which a probability is assigned.
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