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Lesson 4
Solving Random
Variables
Objectives
At the end of this lesson, the learner should be able to
โ— accurately solve possible values of a random
variable; and
โ— correctly solve real-life problems involving the
random variables.
Essential Questions
โ— How do you solve for the possible values of a random
variable?
โ— What is the relationship of probability in solving for
possible values of a random variable?
โ— How are you able to solve problems involving random
variables?
Warm Up!
Before we learn about solving random variables, let us have a
short game where I will ask questions/pick up dares to
chosen students using the online wheel.
(Click on the link to access the online wheel.)
โ€œWheel Decide.โ€ Wheel Decide. Retrieved 24 June 2019 from
https://wheeldecide.com/.
Guide Questions
โ— Do you think the wheel randomly chose names?
โ— Do you have the control to pick your name in the wheel?
โ— What are the chances that your name will be chosen?
Learn about It!
Discrete Random Variable
is a random variable with a finite number of possible values or an infinite number
of values that can be counted
1
Example:
The number of pupils, number of pencils, cards in a standard
deck are examples of a discrete random variable since it can
be counted as 0, 1, 2, 3, and so on. It has an infinite number of
values that can be counted.
Try It!
Example 1:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3).
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
Try It!
Example 1:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3).
Solution:
In a probability distribution for a discrete random variable, the
sum of all the probabilities of the outcomes should be equal to
1. That is ๐‘ƒ ๐‘‹ = 1.
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
Try It!
Example 1:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3).
Solution:
Add all the probabilities and equate it to 1 to find the missing
value.
๐‘ƒ ๐‘‹ = 1 + ๐‘ƒ ๐‘‹ = 2 + ๐‘ƒ ๐‘‹ = 3 + ๐‘ƒ ๐‘‹ = 4 = 1
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
Try It!
Example 1:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3).
Solution:
0.10 + 0.21 + ๐‘ƒ ๐‘‹ = 3 + 0.5 = 1
0.81 + ๐‘ƒ ๐‘‹ = 3 = 1
๐‘ƒ ๐‘‹ = 3 = 1 โˆ’ 0.81
๐‘ƒ ๐‘‹ = 3 = 0.19
Thus, ๐‘ท ๐‘ฟ = ๐Ÿ‘ = ๐ŸŽ. ๐Ÿ๐Ÿ—.
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
Try It!
Example 2:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3).
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 1
10
3
10
2
10
4
10
Try It!
Example 2:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3).
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 1
10
3
10
2
10
4
10
Solution:
Analyze the problem.
The notation ๐‘ƒ(๐‘‹ < 3) means that the random variable ๐‘‹ is less
than 3.
Try It!
Example 2:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3).
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 1
10
3
10
2
10
4
10
Solution:
The random variables take on the values 1, 2, 3, and 4 in which 1
and 2 are less than 3. The probabilities are ๐‘ƒ(๐‘‹ = 1) and ๐‘ƒ(๐‘‹ =
2). Add all the identified probabilities.
๐‘ƒ ๐‘‹ < 3 = ๐‘ƒ ๐‘‹ = 1 + ๐‘ƒ ๐‘‹ = 2
Try It!
Example 2:
Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3).
๐‘ฟ 1 2 3 4
๐‘ท(๐‘ฟ) 1
10
3
10
2
10
4
10
Solution:
๐‘ƒ ๐‘‹ < 3 = ๐‘ƒ ๐‘‹ = 1 + ๐‘ƒ ๐‘‹ = 2
๐‘ƒ ๐‘‹ < 3 =
1
10
+
3
10
=
4
10
or
2
5
Therefore, the probability that the random variable ๐‘‹ is less
than 3 is
๐Ÿ’
๐Ÿ๐ŸŽ
or
๐Ÿ
๐Ÿ“
.
Letโ€™s Practice!
Individual Practice:
1. Below is a valid probability distribution of a random
variable ๐‘‹. Solve for the missing value.
๐‘ฟ 3 5 6 9
๐‘ท(๐‘ฟ) 1
5
๐‘˜ 2
10
2๐‘˜
Letโ€™s Practice!
Individual Practice:
2. Given the probability distribution of a discrete random
variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ โ‰ฅ 78).
๐‘ฟ 77 78 79 80 81
๐‘ท(๐‘ฟ) 0.135 0.292 0.284 0.230 0.059
Letโ€™s Practice!
Group Practice: The class will be divided into 5 groups.
The number of hours that a student spends in studying at
home is a random variable ๐‘‹ given by ๐‘ƒ ๐‘‹ =
๐‘ฅ+2
9
, where
๐‘ฅ = 0, 1, and 2.
a. Construct the probability distribution for the random
variable ๐‘‹.
b. What is the probability that a student spent at least 1 hour
studying at home?
Key Points
Discrete Random Variable
is a random variable with a finite number of possible values or an infinite number
of values that can be counted.
1
Synthesis
โ— How do we solve possible values of a random variable?
โ— How are random variables applicable in our daily lives?
Can you give a specific example?
โ— How do you know the outcome that will most likely to
happen?

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  • 2. Objectives At the end of this lesson, the learner should be able to โ— accurately solve possible values of a random variable; and โ— correctly solve real-life problems involving the random variables.
  • 3. Essential Questions โ— How do you solve for the possible values of a random variable? โ— What is the relationship of probability in solving for possible values of a random variable? โ— How are you able to solve problems involving random variables?
  • 4. Warm Up! Before we learn about solving random variables, let us have a short game where I will ask questions/pick up dares to chosen students using the online wheel. (Click on the link to access the online wheel.) โ€œWheel Decide.โ€ Wheel Decide. Retrieved 24 June 2019 from https://wheeldecide.com/.
  • 5. Guide Questions โ— Do you think the wheel randomly chose names? โ— Do you have the control to pick your name in the wheel? โ— What are the chances that your name will be chosen?
  • 6. Learn about It! Discrete Random Variable is a random variable with a finite number of possible values or an infinite number of values that can be counted 1 Example: The number of pupils, number of pencils, cards in a standard deck are examples of a discrete random variable since it can be counted as 0, 1, 2, 3, and so on. It has an infinite number of values that can be counted.
  • 7. Try It! Example 1: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3). ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
  • 8. Try It! Example 1: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3). Solution: In a probability distribution for a discrete random variable, the sum of all the probabilities of the outcomes should be equal to 1. That is ๐‘ƒ ๐‘‹ = 1. ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
  • 9. Try It! Example 1: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3). Solution: Add all the probabilities and equate it to 1 to find the missing value. ๐‘ƒ ๐‘‹ = 1 + ๐‘ƒ ๐‘‹ = 2 + ๐‘ƒ ๐‘‹ = 3 + ๐‘ƒ ๐‘‹ = 4 = 1 ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
  • 10. Try It! Example 1: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ = 3). Solution: 0.10 + 0.21 + ๐‘ƒ ๐‘‹ = 3 + 0.5 = 1 0.81 + ๐‘ƒ ๐‘‹ = 3 = 1 ๐‘ƒ ๐‘‹ = 3 = 1 โˆ’ 0.81 ๐‘ƒ ๐‘‹ = 3 = 0.19 Thus, ๐‘ท ๐‘ฟ = ๐Ÿ‘ = ๐ŸŽ. ๐Ÿ๐Ÿ—. ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 0.10 0.21 ? 0.5
  • 11. Try It! Example 2: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3). ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 1 10 3 10 2 10 4 10
  • 12. Try It! Example 2: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3). ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 1 10 3 10 2 10 4 10 Solution: Analyze the problem. The notation ๐‘ƒ(๐‘‹ < 3) means that the random variable ๐‘‹ is less than 3.
  • 13. Try It! Example 2: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3). ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 1 10 3 10 2 10 4 10 Solution: The random variables take on the values 1, 2, 3, and 4 in which 1 and 2 are less than 3. The probabilities are ๐‘ƒ(๐‘‹ = 1) and ๐‘ƒ(๐‘‹ = 2). Add all the identified probabilities. ๐‘ƒ ๐‘‹ < 3 = ๐‘ƒ ๐‘‹ = 1 + ๐‘ƒ ๐‘‹ = 2
  • 14. Try It! Example 2: Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ < 3). ๐‘ฟ 1 2 3 4 ๐‘ท(๐‘ฟ) 1 10 3 10 2 10 4 10 Solution: ๐‘ƒ ๐‘‹ < 3 = ๐‘ƒ ๐‘‹ = 1 + ๐‘ƒ ๐‘‹ = 2 ๐‘ƒ ๐‘‹ < 3 = 1 10 + 3 10 = 4 10 or 2 5 Therefore, the probability that the random variable ๐‘‹ is less than 3 is ๐Ÿ’ ๐Ÿ๐ŸŽ or ๐Ÿ ๐Ÿ“ .
  • 15. Letโ€™s Practice! Individual Practice: 1. Below is a valid probability distribution of a random variable ๐‘‹. Solve for the missing value. ๐‘ฟ 3 5 6 9 ๐‘ท(๐‘ฟ) 1 5 ๐‘˜ 2 10 2๐‘˜
  • 16. Letโ€™s Practice! Individual Practice: 2. Given the probability distribution of a discrete random variable ๐‘‹ below, find ๐‘ƒ(๐‘‹ โ‰ฅ 78). ๐‘ฟ 77 78 79 80 81 ๐‘ท(๐‘ฟ) 0.135 0.292 0.284 0.230 0.059
  • 17. Letโ€™s Practice! Group Practice: The class will be divided into 5 groups. The number of hours that a student spends in studying at home is a random variable ๐‘‹ given by ๐‘ƒ ๐‘‹ = ๐‘ฅ+2 9 , where ๐‘ฅ = 0, 1, and 2. a. Construct the probability distribution for the random variable ๐‘‹. b. What is the probability that a student spent at least 1 hour studying at home?
  • 18. Key Points Discrete Random Variable is a random variable with a finite number of possible values or an infinite number of values that can be counted. 1
  • 19. Synthesis โ— How do we solve possible values of a random variable? โ— How are random variables applicable in our daily lives? Can you give a specific example? โ— How do you know the outcome that will most likely to happen?