Question

a company also tries to encourage sales by periodically offering 10% off coupon codes. suppose that we select a day at random. let ( a = ) a coupon code is offered and let ( b = ) the customer makes a purchase. which of the following equalities would indicate that events ( a ) and ( b ) are independent? check all that apply. ( square p(a)=p(b) ) ( square p(a|b)=p(a|b^{c}) ) ( square p(a|b)=p(a) ) ( square p(a|b)=p(b) ) ( square p(a|b^{c})=p(a) )

Explanation:

• For \(P(A|B)=P(A|B^{C})\):
• If \(P(A|B) = P(A|B^{C})\), it means that whether \(B\) occurs or not, the probability of \(A\) is the same. So \(B\) has no impact on \(A\), which is a sign of independence.
• For \(P(A|B)=P(A)\):
• By the definition of conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). If \(P(A|B) = P(A)\), then \(\frac{P(A\cap B)}{P(B)}=P(A)\), which implies \(P(A\cap B)=P(A)P(B)\), the formula for independent events.
• For \(P(A|B^{C})=P(A)\):
• Using the formula \(P(A|B^{C})=\frac{P(A\cap B^{C})}{P(B^{C})}\). If \(P(A|B^{C}) = P(A)\), then \(P(A\cap B^{C})=P(A)P(B^{C})\). Also, since \(P(A)=P(A\cap B)+P(A\cap B^{C})\) and \(P(B^{C}) = 1 - P(B)\), we can show that \(P(A\cap B)=P(A)P(B)\) (using \(P(A\cap B^{C})=P(A)-P(A\cap B)\) and \(P(B^{C})=1 - P(B)\)), which is the condition for independence.
• \(P(A)=P(B)\) just means the probabilities of \(A\) and \(B\) are equal, but it doesn't imply independence. For example, if \(A\) and \(B\) are mutually - exclusive events with \(P(A)=P(B)=\frac{1}{2}\), they are not independent.
• \(P(A|B)=P(B)\) is not a condition for independence. The correct relationship for independence is related to \(P(A)\) and \(P(A\cap B)\)

Answer:

\(\square P(A|B) = P(A|B^{C})\), \(\square P(A|B)=P(A)\), \(\square P(A|B^{C})=P(A)\)