Question

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 and P(B)=0.4.
a. Could it be the case that P( A ∩ B )=0.5, why or why not?

b. From now on, suppose that P( A ∩ B )=0.3 What is the probability that student has one of these two types of cards?

c. What is the probability that the selected student has neither type of card?

d. Describe in terms of A and B the event that the select student has a visa card, but not a mastercard? and then calulate the probability of this event.

e. Calcuate th probability that the selected student has exactly one of the two types of cards?

Answer 1

(a) P( A ∩ B )=0.5 is not possible.

(b) 0.7

(c) 0.3

(d) 0.3

(e) 0.4

Explanation:

Given information: The alphabet A and B represents the following events

A : Individual has a Visa credit card.

B: Individual has a MasterCard.

P(A)= 0.6 and P(B)=0.4.

(a)

We need to check whether P( A ∩ B ) can be 0.5 or not.

`A\cap B\subset A` and
`A\cap B\subset B`

`P(A\cap B)\leq P(A)` and
`P(A\cap B)\leq P(B)`

`P(A\cap B)\leq 0.6` and
`P(A\cap B)\leq 0.4`

From these two inequalities we conclude that

`P(A\cap B)\leq 0.4`

Therefore, P( A ∩ B )=0.5 is not possible.

(b)

Let
`P(A\cap B)=0.3`

We need to find the probability that student has one of these two types of cards.

`P(A\cup B)=P(A)+P(B)-P(A\cap B)`

Substitute the given values.

`P(A\cup B)=0.6+0.4-0.3=0.7`

Therefore the probability that student has one of these two types of cards is 0.7.

(c)

We need to find the probability that the selected student has neither type of card.

`P(A'\cup B')=1-P(A\cup B)`

`P(A'\cup B')=1-0.7=0.3`

Therefore the probability that the selected student has neither type of card is 0.3.

(d)

The event that the select student has a visa card, but not a mastercard is defined as

`A-B`

It can also written as

`A\cap B'`

The probability of this event is

`P(A\cap B')=P(A)-P(A\cap B)`

`P(A\cap B')=0.6-0.3=0.3`

Therefore the probability that the select student has a visa card, but not a mastercard is 0.3.

(e)

We need to find the probability that the selected student has exactly one of the two types of cards.

`P(A\cap B')+P(A\cap B')=P(A\cup B)-P(A\cap B)`

`P(A\cap B')+P(A\cap B')=0.7-0.3`

`P(A\cap B')+P(A\cap B')=0.4`

Therefore the probability that the selected student has exactly one of the two types of cards is 0.4.