SOLUTION: A survey found that 62% of callers in the United States complain about the service they receive from a call center if they suspect that the agent who handled the call is foreign.11

Question 1171069: A survey found that 62% of callers in the United States complain about the service they receive from a call center if they suspect that the agent who handled the call is foreign.11 Given this context, what is the probability that (state your assumptions)
(a) the next three consecutive callers complain about the service provided by a foreign agent?
(b) the next two calls produce a complaint, but not the third?
(c) two out of the next three calls produce a complaint?
(d) none of the next 10 calls produces a complaint?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
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

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