Question

Return to the credit card scenario of Exercise, and let C be the event that the selected student has an American Express card. In addition to\(P\left( A \right) = 0.6\), \(P\left( B \right) = 0.4\), and\(P\left( {A \cap B} \right) = 0.3\), suppose that\(P\left( C \right) = 0.2\), \(P\left( {A \cap C} \right)\; = 0.15\), \(P\left( {B \cap C} \right) = 0.1\), and \(P\left( {A \cap B \cap C} \right) = 0.08\)

a. What is the probability that the selected student has at least one of the three types of cards?

b. What is the probability that the selected student has both a Visa card and a MasterCard but not an American Express card?

c. Calculate and interpret \(P\left( {B|A} \right)\)and also \(P\left( {A|B} \right)\)

d. If we learn that the selected student has an American Express card, what is the probability that she or he also has both a Visa card and a MasterCard?

e. Given that the selected student has an American Express card, what is the probability that she or he has at least one of the other two types of cards?

Short answer

a. The probability that the selected student has at least one of the three types of cards is \(0.73\).

b. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card is \(0.22\).

c. \(P\left( {A|B} \right) = 0.75\) and \(P\left( {A|B} \right) = 0.5\)

d. The probability that she or he also has both a Visa card and a MasterCard is\(0.4\).

e. The probability that she or he has at least one of the other two types of cards is 0.85.

Step-by-step solution

Definition of Probability

Probability is a discipline of mathematics concerned with numerical explanations of the likelihood of an event occurring or the truth of a claim. The probability of an event is a number between \(0\;and{\rm{ }}1\), with zero indicating impossibility and one indicating certainty, broadly speaking.

Calculation for determining probability in part (a)

The chancethat the student has at least one of the three occurrences can also be written as the probability that the student has A, B, or C, which is the unionof the three events.

For every three events, A, B and C, the following equation will be true

\(P(A \cup B \cup C){\rm{ }} = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\)

Using this, we have

\(\begin{array}{c}P(A \cup B \cup C) &=& P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\\ &=& 0.6 + 0.4 + 0.2 - 0.3 - 0.5 - 0.1 + .08\\ &=& 0.73\end{array}\)

Calculation for determining probability in part (b)

The probability that the selected student has both a Visa and a MasterCard but no American Express card can also be written as the probability that the student has A and B but no C, i.e., the intersection of the three events.

We can see from the Venn Diagram below (shaded region) that

\(\begin{array}{c}P\left( {A \cap B \cap {C^\prime }} \right) &=& P(A \cap B) - P(A \cap B \cap C)\\ &=& 0.3 - 0.08\\ &=& 0.22\end{array}\)

Explanation for the determination of probability in part (c)  

Conditional probability of A given that the event B has occurred, for which \(P\left( B \right) > 0\),vis

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\)

for any two event A and B.

Using this,

\(P(B\mid A) = \frac{{P(B \cap A)}}{{P(A)}}\mathop = \limits^{(1)} \frac{{0.3}}{{0.6}} = 0.5,\)

(1): given in the exercise.

Further explanation for the determination of probability in part (c)  

And

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\mathop = \limits^{(1)} \frac{{0.3}}{{0.4}} = 0.75\)

(1): given in the exercise.

Interpretation of the \(P(B\mid A) = 0.5\), When we know a student has a Visa card, the probability that he or she has a MasterCard card will be \(0.5\) (half student who have Visa, have a MasterCard too).

Interpretation of the \(P(B\mid A) = 0.75\), When we know a student has a Visa card, the probability that he or she has a MasterCard card will be \(0.75\)(three-quarters of students with a MasterCard also have a Visa).

Explanation for the determination of probability in part (d)  

In the exercise, we are given probabilities for certain events.

Probability that a student with an American Express card (event C) also possesses a Visa and MasterCard card (\(A \cap B\) because of "and") is

\(\begin{align} P(A\cap B\mid C)&=\frac{P[(A\cap B)\cap C]}{P(C)}......(1) \\ & =\frac{P(A\cap B\cap C)}{P(C)} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=\frac{0.08}{0.2}......(2) \\ &=0.4 \\ \end{align}\)

(1): here we use definition of conditional probability given below, where \(A \cap B\) is a single event,

(2): we are given the probabilities in the exercise.

Conditional probability of A given that the event B has occurred, for which \(P\left( B \right) > 0,\)is

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\)

for any two event A and B.

Explanation for the determination of probability in part (e)  

Probability that an American Express cardholder (event C) also possesses a Visa or MasterCard card (\(A \cup B\)because of "or") is

\(\begin{align} P(A\cup B\mid C)&=\frac{P[(A\cup B)\cap C]}{P(C)}................(1) \\ & =\frac{P[(A\cap C)\cup (B\cap C)]}{P(C)}.................(2) \\ & =\frac{P(A\cap C)+P(B\cap C)-P[(A\cap C)\cap (B\cap C)]}{P(C)}................(3) \\ & =\frac{P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)}{P(C)} \\ & =\frac{0.15+0.1-0.08}{0.2}................(4) \\ & =0.85, \end{align}\)

Further explanation for the determination of probability in part (e)  

(1): here we use definition of conditional probability given below, where \(A \cap B\) is a single event,

(2): this stands for any three events A, B and C,

(3): we use the proposition given below where events \(A \cap B\) and \(B \cap C\) are the single events (first one is \({A_1} = A \cap B\) from the definition, second one is \({B_1} = B \cap C\) from the definition),

(4): the probabilities are given in the exercise.

For every two events \({A_1}\) and \({B_1}\)

\(P\left( {{A_1} \cup {B_1}} \right) = P\left( {{A_1}} \right) + P\left( {{B_1}} \right) - P\left( {{A_1} \cap {B_1}} \right).\)

Conditional probability of A given that the event B has occurred, for which \(P(B) > 0\), is

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\)

for any two events A and B.