Answer:
Step-by-step explanation:
Given that,
Visa card is represented by P(A)
MasterCard is represented by P(B)
P(A)= 0.6
P(A')=0.4
P(B)=0.5
P(B')=0.5
P(A∩B)=0.35
1. P(A U B) =?
P(A U B)= P(A)+P(B)-P(A ∩ B)
P(A U B)=0.6+0.5-0.35
P(A U B)= 0.75
The probability of student that has least one of the cards is 0.75
2. Probability of the neither of the student have the card is given as
P(A U B)'=1-P(A U B)
P(A U B)= 1-0.75
P(A U B)= 0.25
3. Probability of Visa card only,
P(A)= 0.6
P(A) only means students who has visa card but not MasterCard.
P(A) only= P(A) - P(A ∩ B)
P(A) only=0.6-0.35
P(A) only=0.25.
4. Compute the following
a. A ∩ B'
b. A ∪ B'
c. A' ∪ B'
d. A' ∩ B'
e. A' ∩ B
a. A ∩ B'
P(A∩ B') implies that the probability of A without B i.e probability of A only and it has been obtain in question 3.
P(A ∩ B')= P(A-B)=P(A)-P(A∩ B)
P(A∩ B')= 0.6-0.35
P(A∩ B')= 0.25
b. P(A ∪ B')
P(A ∪ B')= P(A)+P(B')-P(A ∩ B')
P(A ∪ B')= 0.6+0.5-0.25
P(A ∪ B')= 0.85
c. P(A' ∪ B')= P(A')+P(B')-P(A' ∩ B')
But using Demorgan theorem
P(A∩B)'=P(A' ∪ B')
P(A∩B)'=1-P(A∩B)
P(A∩B)'=1-0.35
P(A∩B)'=0.65
Then, P(A∩B)'=P(A' ∪ B')= 0.65
d. P( A' ∩ B' )
Using demorgan theorem
P(A U B)'= P(A' ∩ B')
P(A U B)'= 1-P(A U B)
P(A' ∩ B')= 1-0.75
P(A' ∩ B')= 0.25
P(A U B)'= P(A' ∩ B')=0.25
e. P(A' ∩ B)= P(B ∩ A') commutative law
Then, P(B ∩ A') = P(B) only
P(B ∩ A') = P(B) -P(A ∩ B)
P(B ∩ A') =0.5 -0.35
P(B ∩ A') =0.15
P(A' ∩ B)= P(B ∩ A') =0.15
