Question 1171069
You've asked this question before, and I've provided a detailed answer. Let me reiterate the solution with the assumptions clearly stated.

**Assumptions:**

1.  **Independence:** We assume that each caller's complaint is independent of the other callers.
2.  **Constant Probability:** We assume that the probability of a caller complaining about the service provided by a foreign agent is constant at 62% (0.62).

**Given Information:**

* Probability of a complaint (p) = 0.62
* Probability of no complaint (q) = 1 - p = 1 - 0.62 = 0.38

**Calculations:**

**(a) The next three consecutive callers complain.**

* Since the events are independent, we multiply the probabilities:
    * P(complaint, complaint, complaint) = p * p * p = p^3
    * P(3 complaints) = (0.62)^3 = 0.238328
    * Rounded to 4 decimal places: 0.2383

**(b) The next two calls produce a complaint, but not the third.**

* P(complaint, complaint, no complaint) = p * p * q = p^2 * q
    * P(2 complaints, then no complaint) = (0.62)^2 * (0.38) = 0.3844 * 0.38 = 0.146072
    * Rounded to 4 decimal places: 0.1461

**(c) Two out of the next three calls produce a complaint.**

* There are three possible scenarios:
    * complaint, complaint, no complaint (p * p * q)
    * complaint, no complaint, complaint (p * q * p)
    * no complaint, complaint, complaint (q * p * p)
* Each of these scenarios has the same probability: p^2 * q
* Since there are three scenarios, we multiply by 3:
    * P(2 complaints out of 3) = 3 * p^2 * q
    * P(2 complaints out of 3) = 3 * (0.62)^2 * (0.38) = 3 * 0.146072 = 0.438216
    * Rounded to 4 decimal places: 0.4382

**(d) None of the next 10 calls produces a complaint.**

* P(no complaint) = q = 0.38
* Since the events are independent, we multiply the probabilities:
    * P(no complaint for 10 calls) = q^10
    * P(no complaint for 10 calls) = (0.38)^10 = 0.000062788
    * Rounded to 4 decimal places: 0.0001