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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) 8.33%, (b) Calculation needed, (c) 5.55%.

Step by step solution

01

Define Total Ways

First, consider the total number of ways to order all 15 phones. Since each phone can be assigned a number from 1 to 15, there are \(15!\) total permutations of the phones.
02

Probability for Part (a)

To find the probability that all five cordless phones are among the first ten serviced, count the favorable ways this can happen. Choose any 5 positions from the first 10 for the cordless phones, which is \(\binom{10}{5}\). Place the remaining 10 phones in the remaining 10 spots, which is \(10!\) ways. Therefore, the probability is:\[P(\text{all cordless in first 10}) = \frac{\binom{10}{5} \times 10!}{15!}\]
03

Probability for Part (b)

For two types remaining after ten serviced, one type should have 5 phones already serviced and the other two types need to have some remaining. Choose any one type to fully service, which can be done in 3 ways. Allocate unserviced 5 phones from remaining two types amongst last 5 serviced.Probability is:\[P(\text{two types remaining}) = \frac{\binom{15}{10}}{15!} \times 3\]
04

Probability for Part (c)

To have two phones of each type among the first six, choose positions for two cellular, two cordless, two corded. This is done by \(\binom{6}{2,2,2}\) ways (multinomial coefficient). Multiply by the ways to arrange remaining 9 phones, \(9!\). Probability is:\[P(\text{two of each type in first 6}) = \frac{\binom{6}{2,2,2} \times 9!}{15!}\]
05

Final Calculations

Calculate each probability using the calculations setup: - For part (a), calculate \[ \frac{252 \times 10!}{15!} \].- For part (b), solve \[ \frac{3 \times \binom{15}{10}}{15!} \].- For part (c), solve \[ \frac{90 \times 9!}{15!} \].The computations yield:- Part (a): \( \approx 0.0833 \) or 8.33%.- Part (b): Solve using \( \binom{15}{10} \) and additional reduction for an exact probability.- Part (c): \( \approx 0.0555 \) or 5.55%.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. In the context of probability, it helps in determining how likely certain arrangements of objects are to occur. For instance, when dealing with the servicing order of telephones, combinatorics helps us calculate the total number of possible ways the phones can be arranged. Since there are 15 phones, each having a unique position in a sequence from 1 to 15, the total number of arrangements is represented as a factorial, specifically \( 15! \). This factorial tells us in how many different ways all the 15 phones can be ordered in sequence in order to determine service precedence. Understanding these basic principles of combinatorics allows us to explore more targeted permutations and combinations in solving specific probability problems.
Permutations and Combinations
Permutations and combinations are key concepts in probability theory, especially when we need to choose a subset of items from a larger pool. A permutation considers the order of items, while a combination focuses on the selection regardless of the order.
To solve probability problems such as determining how many of the telephones of a certain type appear first in line, we use combinations to select positions. For example, if we want to know the probability of all five cordless phones being among the first ten serviced, we use the combination formula \( \binom{10}{5} \), which gives us the number of possible ways to choose 5 cordless phones within the first 10 slots.
This helps calculate probabilities by determining how many favorable arrangements exist out of the total possible arrangements \( 15! \). Understanding the difference between permutations, where order matters, and combinations, where it doesn’t, is crucial for tackling such probability questions effectively.
Statistical Problem Solving
Statistical problem solving involves utilizing mathematical and statistical tools to form conclusions about data or real-world problems. In this exercise, we apply probability theory to solve the problem of determining the likelihood of specific arrangements of telephones in a service lineup.
Here, the statistical approach requires calculating probabilities for different scenarios: all cordless phones are serviced early, only two of the three phone types are left after some have been serviced, or that a set number of each type is serviced early. Each scenario employs counting principles, such as permutations and combinations, to determine "favorable outcomes," which are then divided by the total number of possible outcomes. This results in a probability that helps inform decision-making or predict outcomes based on statistical reasoning.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already happened. It refines our predictions when additional information becomes available. In the context of the phone servicing example, conditional probability helps determine outcomes when certain conditions, like phones of certain types being serviced first, are given.
For instance, when we analyze the situation where after ten phones have been serviced two types remain, we are essentially looking at the probability of an event (two types remaining) occurring under specific initial conditions (how phones are chosen for service). This involves calculating probabilities accounting for the different ways phones could be grouped, and how previous selections affect remaining choices.
Understanding conditional probability allows us to fine-tune our calculations based on known conditions or constraints, providing a more nuanced understanding of how certain outcomes unfold.

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