SOLUTION: 2. This problem is about a fast-food business. As it turns out, 60% of their business is via the drive-through window, and 40% is counter-service. Data from the registers reveal th

Algebra ->  Probability-and-statistics -> SOLUTION: 2. This problem is about a fast-food business. As it turns out, 60% of their business is via the drive-through window, and 40% is counter-service. Data from the registers reveal th      Log On


   

Question 1170934: 2. This problem is about a fast-food business. As it turns out, 60% of their business is via the drive-through window, and 40% is counter-service. Data from the registers reveal that 70% of the transactions at the drive-through are via credit/debit cards, while 80% of the transactions at the counter are via credit/debit cards. (The other transactions are via cash.) BWhat is the probability? A.that a random transaction uses cash? B.that a random transaction uses credit cards? C.that a transaction was a credit-card transaction and at the drive-through window? D.that a transaction was a cash transaction and was at the drive-through window? E.that a random transaction was at the counter? F.that a random transaction was at the drive-through window?
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Let's break down this problem step-by-step using probability notation.
**Define Events:**
* D: Drive-through transaction
* C: Counter-service transaction
* Cr: Credit/debit card transaction
* Ca: Cash transaction
**Given Probabilities:**
* P(D) = 0.60
* P(C) = 0.40
* P(Cr | D) = 0.70 (Probability of credit card given drive-through)
* P(Ca | D) = 1 - 0.70 = 0.30 (Probability of cash given drive-through)
* P(Cr | C) = 0.80 (Probability of credit card given counter)
* P(Ca | C) = 1 - 0.80 = 0.20 (Probability of cash given counter)
**Calculations:**
**A. Probability of a random transaction using cash (P(Ca))**
* P(Ca) = P(Ca | D) * P(D) + P(Ca | C) * P(C)
* P(Ca) = (0.30 * 0.60) + (0.20 * 0.40)
* P(Ca) = 0.18 + 0.08 = 0.26
**B. Probability of a random transaction using credit cards (P(Cr))**
* P(Cr) = P(Cr | D) * P(D) + P(Cr | C) * P(C)
* P(Cr) = (0.70 * 0.60) + (0.80 * 0.40)
* P(Cr) = 0.42 + 0.32 = 0.74
**C. Probability of a credit-card transaction at the drive-through (P(Cr ∩ D))**
* P(Cr ∩ D) = P(Cr | D) * P(D)
* P(Cr ∩ D) = 0.70 * 0.60 = 0.42
**D. Probability of a cash transaction at the drive-through (P(Ca ∩ D))**
* P(Ca ∩ D) = P(Ca | D) * P(D)
* P(Ca ∩ D) = 0.30 * 0.60 = 0.18
**E. Probability of a random transaction at the counter (P(C))**
* P(C) = 0.40 (given)
**F. Probability of a random transaction at the drive-through (P(D))**
* P(D) = 0.60 (given)
**Answers:**
* A. P(Ca) = 0.26
* B. P(Cr) = 0.74
* C. P(Cr ∩ D) = 0.42
* D. P(Ca ∩ D) = 0.18
* E. P(C) = 0.40
* F. P(D) = 0.60