91Ó°ÊÓ

Event A represents that the selected student has a Visa card.

Event B represents the selected student who has a Master card.

The probability of event A is \(P\left( A \right) = 0.6\).

The probability of event B is \(P\left( B \right) = 0.4\).

a.

Since \(A \cap B\) is contained in B, this implies that \(P\left( {A \cap B} \right) \le P\left( B \right)\).

This inequality violates as 0.5>0.4.

Therefore, this is not possible.

b.

The probability is \(P\left( {A \cap B} \right) = 0.3\).

The probability that the selected student has at least one of these two types of cards is computed as,

\(\begin{aligned}P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\ &= 0.6 + 0.4 - 0.3\\ &= 0.7\end{aligned}\)

Therefore, the probability that the selected student has at least one of these two types of cards is 0.7.

c.

The probability that the selected student has neither type of card is computed as,

\(\begin{aligned}1 - P\left( {A \cup B} \right) &= 1 - 0.7\\ &= 0.3\end{aligned}\)

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

d.

The event that the selected student has a Visa card but not a MasterCard is represented as \(A \cap B'\).

The probability of the above event is computed as,

\(\begin{aligned}P\left( {A \cap B'} \right) &= P\left( A \right) - P\left( {A \cap B} \right)\\ &= 0.6 - 0.3\\ &= 0.3\end{aligned}\)

Therefore, the probability of the provided event is 0.3.

e.

The probability that the selected student has exactly one of the two types of cards is computed as,

\(\begin{aligned}P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) &= \left( {P\left( A \right) - P\left( {A \cap B} \right)} \right) + \left( {P\left( B \right) - P\left( {A \cap B} \right)} \right)\\ &= \left( {0.6 - 0.3} \right) + \left( {0.4 - 0.3} \right)\\ &= 0.3 + 0.1\\ &= 0.4\end{aligned}\)

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