Probability of Either Event A or B happening, or Both happening

Use a Venn diagram to prove that the probability of either event A or B occurring (A and B are not mutually exclusive) is given by:

P(A∪B)=P(A)+P(B)-P(A∩B)
solution:

Q1. A group of learners is given the following event sets:

The sample space can be described as {n∶n ϵ Z,1≤n≤6}
They are asked to calculate the value of P(A∩B). They get stuck, and you offer to calculate it for them. Give your answer as a decimal number, rounded to two decimal value.
solution:

P(A∪B)=P(A)+P(B)-P(A∩B)
Make P(A∩B) the subject, and we get:

P(A∩B)=P(A)+P(B)-P(A∪B)
Identify variables needed:

Calculate P(A∩B):

P(A∩B)=P(A)+P(B)-P(A∪B)
=⅓+½-⅔=⅙=0.17
Therefore, the value of P(A∩B) is 0.17.

Q2. A group of learners is given the following event sets:

The sample space can be described as {n∶n ϵ Z,1≤n≤6}.
They are asked to calculate the value of P(A∪B). They get stuck, and you offer to calculate it for them. Give your answer as a decimal number, rounded to two decimal places.
solution:

P(A∪B)=P(A)+P(B)-P(A∩B)
Identify variables needed:

Calculate P(A∪B):

P(A∪B)=P(A)+P(B)-P(A∩B)
=⅔+⅙-0=⅚=0.83
Therefore, the value of P(A∪B) is 0.83.

Q3. In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is
selected at random from the class, find the probability that he will be studying Mathematics or Biology.
solution:
Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology.

P(A)=40%=0.4,P(B)=30%=0.3,P(A and B)=0.1
We know that

P(A or B)=P(A)+P(B)-P(A and B)
P(A or B)=0.4+0.3-0.1=0.6
Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6.

Q4. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both?
solution:
Let A and B be the events of passing first and second examinations respectively.
Accordingly, P(A)=0.8,P(B)=0.7,P(A or B)=0.95
We know that

P(A or B)=P(A)+P(B)-P(A and B)
0.95=0.8+0.7-P(A and B)
P(A and B)=1.5-0.95=0.55
Thus, the probability of passing both the examinations is 0.55.

Q5. A card is selected at random from a deck of 52 cards. Find the probability that it is a 7 or a club.
solution:
Let A = the event of getting a 7; then P(A)=4/52 since there are four 7s. Let B = the event of getting a club; then P(B)=13/52 since there are 13 clubs.
Since there is one card that is both a 7 and a club (i.e., the 7 of clubs),

P(A or B)=P(A)+P(B)-P(A and B)

Q6. A die is rolled. Find the probability of getting an odd number or a number less than 5.
solution:
Let A=an odd number; then P(A)=3/6 since there are three odd numbers →namely 1,3, and 5. Let B=a number less than 5; then P(B)=4/6 since there are four numbers less than 5 →namely 1, 2, 3, and 4. Let P(A and B)=4/6. Sincethere are two odd numbers less than 5 →1 and 3. Hence,

P(A or B)=P(A)+P(B)-P(A and B)

Q7. A natural number is chosen out of the first fifty natural numbers. What is the probability that the chosen numbers is a multiple of 3 or of 5?
solution:

n(S)=50
If A is the event that the chosen number is multiple of 3.
Let A={3,6,9,…,48}

n(A)=16
P(A)=16/50=8/25
If B is the event that the chosen number is multiple of 5.

B={5,10,15,…,50}
n(B)=10
P(B)=10/50=⅕
A∩B={15,30,45}→n(A∩B)=3
Since A and B are not mutually exclusive

Q8. At a sales meeting, there were 15 men and 24 women. Eight of the men and 10 of the women were selected to sell a new product. If a sales person were selected at random, find the probability that the sales person was a female or a person who was selected to sell the new product.
solution:
There were 15+24=39 people at the meeting. Since there are 24 women,P(women)=24/39, and since 8+10=18 of the salespersons were selected tosell the new product, P(person selected to sell new product). itk.The number of sales people who are female and who were selected to sell the newproduct Is 10. Hence,

P(female or sales person)
=P(female)+P(sales person)-P(female and sales person)

Q9. In year 7 at Mt Random High School, every student must do art or music.In a group of 100 students surveyed, 47 do music and 59 do art. If one student is chosen at random from year 7, find the probability that this student does
(a) both art and music
(b) only art
(c) only music
solution:

47+59=106
But there are only 100 students!
This means 6 students have been counted twice.
i.e. 6 students do both art and music.
Students doing music only: 47-6=41
Students doing art only: 59-6=53
A Venn diagram shows this information.

(a) P(both) =6/100=3/50,
(b) P(art only) =53/100
(c) P(music only) =41/100

Q10: Reasoning with Venn diagrams
In a survey 70 people were questioned about which product they use: A or B or both. The report of the survey shows that 25 people use product A, 35 people use product B and 15 people use neither. Use a Venn diagram to work out how many people:
1. use product A only
2. use product B only
3. use both product A and product B
solution:
Step 1: Summarise the sizes of the sample space, the event sets, their union and their intersection
We are told that 70 people were questioned, so the size of the sample space is n(S)=70. We are told that 25 people use product A, so n(A)=25.
We are told that 35 people use product B, so n(B)=35.
We are told that 15 people use neither product. This means that 70-15=55 people use at least one of the two products, so n(A∪B)=55.
We are not told how many people use both products, so we have to work out the size of the intersection, A∩B, by using the identity for the union of two events:

Step 2: Determine whether the events are mutually exclusive
Since the intersection of the events, A∩B , is not empty, the events are not mutually exclusive. This means that their circles should overlap in the Venn diagram.
Step 3: Draw the Venn diagram and fill in the numbers