91Ó°ÊÓ

To evaluate the technical support from a computer manufacturer, the number of rings before a call is answered by a service representative is tracked. Historically, \(70 \%\) of the calls are answered in two rings or less, \(25 \%\) are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume that the calls are independent. (a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more? (b) What is the probability that all 10 calls are answered in four rings or less? (c) What is the expected number of calls answered in four rings or less? (d) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less? (e) What is the conditional expected number of calls requiring five rings or more given that eight calls are answered in two rings or less? (f) Are the number of calls answered in two rings or less and the number of calls requiring five rings or more independent random variables?

Step by step solution

01

Understanding the problem

We have three independent events for call answering times: - Calls answered in two rings or less ( 0.7). - Calls answered in three or four rings ( 0.25). - Calls needing five or more rings ( 0.05). Let us denote these events as events A, B, and C, respectively. We need to use this information to solve multiple sub-questions considering 10 independent call instances.

02

Part (a) Probability Calculation

To calculate the probability that precisely 8 calls are event A, 1 call is event B, and 1 call is event C, we use the multinomial distribution formula:\[ P(X_A = 8, X_B = 1, X_C = 1) = \frac{10!}{8!1!1!} (0.7)^8 (0.25)^1 (0.05)^1 \]Calculating:- \(\frac{10!}{8!1!1!} = 45\)- \((0.7)^8 \approx 0.057648009\)- \((0.25)^1 = 0.25\)- \((0.05)^1 = 0.05\)Thus, \( P \approx 45 \times 0.057648009 \times 0.25 \times 0.05 \approx 0.032\).

03

Part (b) Probability Calculation

To find the probability that all 10 calls are answered in 4 rings or less, the calls have outcomes in event A or B. So, the probability of these outcomes is 0.7 + 0.25 = 0.95 for each call. Using the binomial distribution formula, we have:\[ P(X_A + X_B = 10) = (0.95)^{10} \]Calculating:- \((0.95)^{10} \approx 0.59874\).Thus, \( P \approx 0.599\).

04

Part (c) Expectation Calculation

The expected number of calls answered in four rings or less is calculated as the sum of calls in events A and B:\[ E(X_A + X_B) = n(p_A + p_B) = 10 \times 0.95 = 9.5 \]Thus, the expected number is 9.5.

05

Part (d) Conditional Distribution

To solve for the conditional distribution of calls requiring 5 or more rings given that 8 are answered in 2 or less, we need:Based on the question's condition, 8 calls are A and the rest are divided between B and C (2 calls).Let \(Y = X_C \). Then, \(Y = 10 - 8 - X_B = 2 - X_B\).So \(Y\) must be 0 or 1 or 2 based on whether 2, 1, or 0 calls are B.

06

Part (e) Conditional Expectation

Given 8 calls are A, find the expected number of calls C.If 8 are A, focus on the remaining 2 calls that are B and C. With the same logic:- E[C] = 2 * proportion of C among B and C- Probability for C: 0.05/(0.25+0.05)=0.05/0.3The expected number is then, over 2 trials:\[ E(C | \text{8 A}) = 2 \times \left(\frac{1}{6}\right) \approx 0.33\].

07

Part (f) Independence Check

To determine independence, check if one variable's distribution is unaffected by fixing one value of another. Here, knowing the count of A influences B and C because the structural number of problems is fixed, showing they're not independent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multinomial Distribution

The multinomial distribution is an essential concept in probability theory, especially when dealing with experiments where each trial has more than two possible outcomes. In simpler terms, it's a generalization of the binomial distribution to more than two categories. For example, in the exercise, we are considering call responses that can be categorized into three outcomes: answered in two rings or less, answered in three or four rings, and answered with five or more rings.

The probability of observing a specific combination of outcomes, given the number of trials and the probabilities of each outcome, can be determined using the multinomial distribution formula:

• If we denote the probabilities of each event as follows:
  • Event A (two rings or less): 0.7
  • Event B (three or four rings): 0.25
  • Event C (five or more rings): 0.05
  The formula is:\[ P(X_A = x, X_B = y, X_C = z) = \frac{n!}{x!y!z!} (p_A)^x (p_B)^y (p_C)^z \]In this scenario, it was used to find the probability of eight calls being answered in two rings or less, one call in three to four rings, and one call requiring five or more rings.

This is commonly useful when events are independent and we have multiple categories to account for. Understanding multinomial distributions empowers us to analyze complex scenarios involving multiple outcomes.

Binomial Distribution

The binomial distribution is a cornerstone of probability theory, often employed when there are exactly two possible outcomes in a trial, referred to as "success" and "failure". Each trial is independent of the others.

In the context of our exercise, the situation in part (b)—where we want to know the probability that all ten calls are answered in four rings or less—is an example of a binomial distribution problem. Here, the combined probability of either a call being answered in two rings or less or in three to four rings is \[ p = 0.7 + 0.25 = 0.95 \].

We use the binomial formula:

• \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
• Where:
  • \( n \) is the number of trials (10 calls).
  • \( k \) is the number of successful trials (all 10 calls in this case).

Ultimately, the binomial distribution allows us to calculate the likelihood of a specific number of successes from a fixed number of identical, independent experiments. It's extremely practical for scenarios like call answer times where each trial has two potential outcomes.

Conditional Probability

Conditional probability explores the likelihood of an event occurring given that another event has already taken place. It's a pivotal concept in probability theory, allowing for refined probability calculations when additional information is available.

In the exercise, part (d) asks about the conditional distribution of the number of calls requiring five rings or more, given that eight calls are already answered in two rings or less. To deal with such problems, we rely on conditional probability principles.

The general formula for conditional probability is:

• \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]

Here, the focus is on how the remaining calls can be distributed between the two other categories, given the constraint on calls in two rings or less. Recognizing and applying conditional probability appropriately can simplify analyses involving multiple related events.

Mathematical Expectation

Mathematical expectation, often referred to as the expected value, is a key concept in probability that represents the average or mean value of a random variable over a large number of trials. It provides a measure of the "central tendency" of the probability distribution of the variable.

For our exercise, part (c) and (e) involve calculating the expected number of calls answered in certain ways. The expectation with respect to outcomes can be calculated using:

• \[ E(X) = n \cdot p \]
• Where \( n \) is the number of trials and \( p \) is the probability of the event.

This formula gives the long-run average number of successes (in this context, calls answered in certain time frames). Lines like these enable us not only to anticipate quantitative outcomes efficiently but to also plan for them, maximizing resources or adjusting strategies in settings ranging from customer support operations to complex financial models.

Mathematical expectation plays a vital role, offering a statistical baseline around which real-world outcomes fluctuate.