Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Visa credit card and B be the analogous event for a MasterCard. Suppose that P(A) = 0.6, P(B) = 0.5, and P(A n B) = 0.15.

(a) Compute the probability that the selected individual has at least one of the two types of cards (i. e., the probability of the event A ? B).

(b) What is the probability that the selected individual has neither type of card?

c) Describe, in terms of A and B, the event that the selected student has a Visa card but not a MasterCard.
A' n B
A' n B'
A' ? B'
A ? B'
A n B'

Calculate the probability of this event.

a) 0.95

b) 0.05

c) (A∩B')

0.45

Step-by-step explanation:

We know that P(A)=0.6, P(B)=0.5 and P(A∩B)=0.15.

a)

P( At least one of two types of cards)=P(A∪B)=?

P(A∪B)=P(A)+P(B)-P(A∩B)=0.6+0.5-0.15=0.95

Thus, the probability that the selected individual has at least one of the two types of cards is 0.95.

b)

P(neither type of card)=P[(A∪B)']

P[(A∪B)']=1-P(A∪B)=1-0.95=0.05

Thus, the probability that the selected individual has neither type of card is 0.05.

c)

As event A denote that selected individual has a Visa card and event B denote that selected individual has a Master card, so the event that the selected student has a visa card but not a Master card can be denoted as (A∩B').

P(A∩B')=P(A)-P(A∩B)=0.6-0.15=0.45

Thus, the probability that the selected student has a Visa card but not a MasterCard is 0.45.

correct

step-by-step explanation:

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons