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

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