Question 1170934
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