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Chapter 2: Problem 39

Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers \(1,2, \ldots, 15\) to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced? c. What is the probability that two phones of each type are among the first six serviced?

Short Answer

Expert verified
a. \( \approx 0.0839 \), b. \( \approx 0.251 \), c. \( \approx 0.1998 \)

Step by step solution

01

Understanding the Setup

We have 15 phones: 5 cellular, 5 cordless, and 5 corded. They are randomly assigned numbers to determine the service order. We need to find probabilities for certain conditions involving the servicing order.
02

Probability All Cordless Phones Are in the First Ten

We have 5 cordless phones out of 15, and we're interested in them being in the first 10 serviced. First, pick 5 positions out of the first 10 for cordless phones. This is represented by the combination \(\binom{10}{5}\). For the remaining 10 phones (after choosing cordless), assign any positions from the remaining 5 available positions among the first 10. Total possible ways to choose first 10 out of 15 is \(\binom{15}{10}\).Thus, the probability can be calculated as:\[P(\text{all cordless in first 10}) = \frac{\binom{10}{5} \times \binom{5}{5}}{\binom{15}{10}}= \frac{252}{3003}\approx 0.0839\]
03

Probability Only Two of Three Types Remain After Ten

After servicing 10 phones, 5 remain. We want only two types to remain. Consider partitioning the phones into three possible distributions among two types: (5,0,0),(0,5,0),(0,0,5) respectively for types (e.g., cordless vs others). Choose which type to be exhausted and which two types remain; similar operations apply to select phones to fit.Using symmetry and consistency, which involves arranging and grouping possibilities relevant to each context, determine:\[P(\text{only two types remain}) = \text{ } 3 \times \left(\frac{\binom{5}{5} \times \binom{10}{5}}{\binom{15}{10}} \right) \approx \frac{3 \cdot 252}{3003} \approx 0.251\]
04

Probability Two Phones of Each Type Among First Six

Select 2 phones from each type to be among the first 6. The number of ways to do this is given by the product of combinations for each type: \(\binom{5}{2} \cdot \binom{5}{2} \cdot \binom{5}{2}\). Total possible configurations for first 6 chosen from 15 is \(\binom{15}{6}\).Therefore, the probability is:\[P(\text{two of each type in first 6}) = \frac{\binom{5}{2}^3}{\binom{15}{6}} = \frac{10 \cdot 10 \cdot 10}{5005} = \frac{1000}{5005} \approx 0.1998\]

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Key Concepts

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

Combinatorial Probability
A cornerstone of understanding probability in engineering statistics involves combinatorial probability. It's about calculating the number of ways to select items from a set. This ability is crucial in scenarios like determining the order of servicing telephones, where items are selected from distinct groups.

In the context of our exercise, combinatorial probability helps us decide how to arrange telephone types within service orders. We use combinations because the order within each chosen set doesn't matter, only the selection itself.
  • The combination formula is denoted as \( \binom{n}{k} \), which calculates how many ways we can choose \( k \) items from \( n \) without regard to order.
  • For example, \( \binom{10}{5} \) helps us determine how many ways we can pick 5 places for cordless phones within the first 10 service slots.
This method ensures we can accurately quantify the likelihood of various servicing arrangements, painting clearer pictures of each probability scenario.
Telephone Service Orders
In order to determine probability, it's important to understand the service order structure. Our exercise deals with how telephones are randomly assigned positions for servicing in a sequence from 1 to 15.

This setup helps us explore several outcomes:
  • For instance, determining if all of one type, like cordless phones, will appear early in the sequence among the first 10 serviced.
  • Alternatively, it helps us determine combinations where only specific phone types remain after servicing a group of 10.
Thinking about service orders involves both evaluating what sequences are possible and what constraints exist within those sequences. This understanding ensures that probabilities derived are accurate, precise, and reflect real-world scenarios.
Types of Phones in Probability Problems
When solving probability problems involving telephones, it's vital to recognize the distinct groups of phone types and their distributions. Typically, as noted in our exercise, the phones are divided into three types: cellular, cordless, and corded.

In practical scenarios:
  • Different probabilities are associated with whether a certain type like cordless ends up first, or if specific types remain after servicing a segment of phones.
  • This detail helps define the probability distribution across the phone types, contributing to our overall calculation.
Understanding these classifications is critical as they form the basis for applying combinatorial calculations and identifying the correct logic path for each probability question, ensuring our solutions are both mathematically sound and representative of the problem's context.

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Most popular questions from this chapter

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