Problem

# Return to the credit card scenario of Exercise 12 (Section 2.2), where A = {Visa}, B =...

Return to the credit card scenario of Exercise 12 (Section 2.2), where A = {Visa}, B = { MasterCard}, P(A) = .5, P(B) = .4, and P(A ∪B) = .25. Calculate and interpret each of the following probabilities (a Venn diagram might help).

a. P(B|A)

b. P(B ′|A).

c. P(A|B)

d. P(A ′|B)

e. Given that the selected individual has at least one card, what is the probability that he or she has a Visa card?

#### Step-by-Step Solution

Solution 1

(a)

Use the definition of conditional probability along with the given values to find $$P(B \mid A)$$

\begin{aligned} P(B \mid A) &=\frac{P(A \cap B)}{P(A)} \\ &=\frac{0.25}{0.5} \\ &=0.5 \end{aligned}

Therefore, $$P(B \mid A)=0.5$$

(b)

Let $$S$$ denote the entire sample space. Using the distributive law for sets, we have that

\begin{aligned} \left(A \cap B^{\prime}\right) \cup(A \cap B) &=A \cap\left(B^{\prime} \cup B\right) \\ &=A \cap S \\ &=A \end{aligned}

Furthermore, $$A \cap B^{\prime}$$ and $$A \cap B$$ are disjoint events, due to the fact that $$B$$ and $$B^{\prime}$$ are disjoint. The third axiom of probability states that that the probability of the union of disjoint events is the sum of the individual probabilities. Thus

$$P\left(A \cap B^{\prime}\right)+P(A \cap B)=P(A)$$

Rearranging terms and substitute in for known values to calculate $$P\left(A \cap B^{\prime}\right)$$.

\begin{aligned} P\left(A \cap B^{\prime}\right) &=P(A)-P(A \cap B) \\ &=(0.5)-(0.25) \\ &=0.25 \end{aligned}

Use the calculated value of $$P\left(A \cap B^{\prime}\right)$$ and the given value for $$P(A)$$ to compute $$P\left(B^{\prime} \mid A\right)$$.

\begin{aligned} P\left(B^{\prime} \mid A\right) &=\frac{P\left(A \cap B^{\prime}\right)}{P(A)} \\ &=\frac{0.25}{0.5} \\ &=0.5 \end{aligned}

Therefore, $$P\left(B^{\prime} \mid A\right)=0.5$$.

(c)

Use the definition of conditional probability along with the given values to find $$P(A \mid B)$$.

\begin{aligned} P(A \mid B) &=\frac{P(A \cap B)}{P(B)} \\ &=\frac{0.25}{0.4} \\ &=0.625 \end{aligned}

Therefore, $$P(A \mid B)=0.625$$.

(d)

Begin by writing the definition of the conditional probability $$P\left(A^{\prime} \mid B\right)$$.

$$P\left(A^{\prime} \mid B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}$$

The value of $$P(B)$$ is given, so it is only necessary to calculate $$P\left(A^{\prime} \cap B\right)$$.

Let $$S$$ denote the entire sample space. Using the distributive law for sets, we have that

\begin{aligned} \left(A^{\prime} \cap B\right) \cup(A \cap B) &=B \cap\left(A^{\prime} \cup A\right) \\ &=B \cap S \\ &=B \end{aligned}

Furthermore, $$A^{\prime} \cap B$$ and $$A \cap B$$ are disjoint events, due to the fact that $$A$$ and $$A^{\prime}$$ are disjoint. The third axiom of probability states that that the probability of the union of disjoint events is the sum of the individual probabilities. Thus

$$P\left(A^{\prime} \cap B\right)+P(A \cap B)=P(B)$$

Rearrange terms and substitute for known values to calculate $$P\left(A \cap B^{\prime}\right)$$.

\begin{aligned} P\left(A^{\prime} \cap B\right) &=P(B)-P(A \cap B) \\ &=(0.4)-(0.25) \\ &=0.15 \end{aligned}

Use the calculated value of $$P\left(A^{\prime} \cap B\right)$$ and the given value for $$P(B)$$ to compute $$P\left(A^{\prime} \mid B\right)$$.

\begin{aligned} P\left(A^{\prime} \mid B\right) &=\frac{P\left(A^{\prime} \cap B\right)}{P(B)} \\ &=\frac{0.15}{0.4} \\ &=0.375 \end{aligned}

Therefore, $$P\left(A^{\prime} \mid B\right)=0.375$$

(e)

The probability to be calculated is $$P(A \mid A \cup B)$$. Begin by writing the definition of this conditional probability and simplifying.

\begin{aligned} P(A \mid A \cup B) &=\frac{P[A \cap(A \cup B)]}{P(A \cup B)} \\ &=\frac{P(A)}{P(A \bigcup B)} \end{aligned}

The value of $$P(A)$$ is given, so it is only necessary to calculate $$P(A \cup B)$$.

Use the formula for the probability of the union of two general events to compute $$P(A \cup B)$$.

\begin{aligned} P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ &=(0.5)+(0.4)-(0.25) \\ &=0.65 \end{aligned}

Use the calculated value of $$P(A \bigcup B)$$ and the given value for $$P(A)$$ to compute $$P(A \mid A \cup B)$$

\begin{aligned} P(A \mid A \cup B) &=\frac{P(A)}{P(A \cup B)} \\ &=\frac{0.5}{0.65} \\ &=0.769 \end{aligned}

Therefore, $$P(A \mid A \cup B)=0.769$$.