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?

(a) P( A ∩ B )=0.5 is not possible.

(b) 0.7

(c) 0.3

(d) 0.3

(e) 0.4

Step-by-step 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.

and

and

and

From these two inequalities we conclude that

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

(b)

Let

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

Substitute the given values.

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.

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

It can also written as

The probability of this event is

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.

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

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.