Answer:
We have that
.
So
![0.3 = b + (A \cap B)](https://tex.z-dn.net/?f=0.3%20%3D%20b%20%2B%20%28A%20%5Ccap%20B%29)
However, b is a probability, which means that it cannot be negative. So no, P(A ∩ B) cannot be 0.5. It can, at most, be 0.3.
Step-by-step explanation:
Event A:
Probability that a student has a Visa card.
Event B:
Probability that the student has a MasterCard.
We have that:
![A = a + (A \cap B)](https://tex.z-dn.net/?f=A%20%3D%20a%20%2B%20%28A%20%5Ccap%20B%29)
In which a is the probability that a student has a Visa card but not a MasterCard and
is the probability that a student has both these cards.
By the same logic, we have that:
![B = b + (A \cap B)](https://tex.z-dn.net/?f=B%20%3D%20b%20%2B%20%28A%20%5Ccap%20B%29)
In this problem, we have that:
![A = 0.6, B = 0.3](https://tex.z-dn.net/?f=A%20%3D%200.6%2C%20B%20%3D%200.3)
(a) Could it be the case that P(A ∩ B) = 0.5?
We have that
.
So
![0.3 = b + (A \cap B)](https://tex.z-dn.net/?f=0.3%20%3D%20b%20%2B%20%28A%20%5Ccap%20B%29)
However, b is a probability, which means that it cannot be negative. So no, P(A ∩ B) cannot be 0.5. It can, at most, be 0.3.