Question

Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual

Answer 1

Answer:

(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.

$A\cap B\subset A$ 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.