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Copykwik has four photocopy machines: \(A, B, C\), and \(D .\) The probability that a given machine will break down on a particular day is \(P(A)=\frac{1}{50} \quad P(B)=\frac{1}{60} \quad P(C)=\frac{1}{75} \quad P(D)=\frac{1}{40}\) Assuming independence, what is the probability on a particular day that a. All four machines will break down? b. None of the machines will break down?

Short Answer

Expert verified
a. The probability of all four machines breaking down on the same day is \(\frac{1}{9{,}000{,}000}\). b. The probability of none of the machines breaking down on the same day is \(\frac{167{,}965}{180{,}000}\).

Step by step solution

01

(a) Calculate the probability of all four machines breaking down

To find the probability of all four machines breaking down at the same time, we need to multiply the individual probabilities, as we assume independence. \[P(A \cap B \cap C \cap D) = P(A) \cdot P(B) \cdot P(C) \cdot P(D)\] Plug in the given probabilities: \[P(A \cap B \cap C \cap D) = \frac{1}{50} \cdot \frac{1}{60} \cdot \frac{1}{75} \cdot \frac{1}{40}\]
02

Simplify the result

Multiply the fractions to obtain the final result: \[P(A \cap B \cap C \cap D) = \frac{1}{9{,}000{,}000}\] So the probability of all four machines breaking down on the same day is \(\frac{1}{9{,}000{,}000}\).
03

(b) Calculate the probability of each machine not breaking down

To find the probability of each machine not breaking down, we need to subtract the individual probabilities from 1: \[P(A') = 1 - P(A) = 1 - \frac{1}{50} = \frac{49}{50}\] \[P(B') = 1 - P(B) = 1 - \frac{1}{60} = \frac{59}{60}\] \[P(C') = 1 - P(C) = 1 - \frac{1}{75} = \frac{74}{75}\] \[P(D') = 1 - P(D) = 1 - \frac{1}{40} = \frac{39}{40}\]
04

Calculate the probability of none of the machines breaking down

Now, we need to multiply the probabilities of each machine not breaking down, since we assume independence: \[P(A' \cap B' \cap C' \cap D') = P(A') \cdot P(B') \cdot P(C') \cdot P(D')\] Plug in the calculated probabilities: \[P(A' \cap B' \cap C' \cap D') = \frac{49}{50} \cdot \frac{59}{60} \cdot \frac{74}{75} \cdot \frac{39}{40}\]
05

Simplify the result

Multiply the fractions to obtain the final result: \[P(A' \cap B' \cap C' \cap D') = \frac{167{,}965}{180{,}000}\] So the probability of none of the machines breaking down on the same day is \(\frac{167{,}965}{180{,}000}\).

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Key Concepts

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

Independent Events
Independent events are a fundamental concept in probability. When two or more events are independent, it means that the outcome of one event does not affect the outcome of the other(s). In other words, the events occur without influencing each other.
To determine if events are independent, consider this: the probability of the events happening together can be found by multiplying their individual probabilities.
  • For example, if the probability of event A occurring is \(P(A)\) and the probability of event B occurring is \(P(B)\), the probability of both events occurring is \(P(A) \times P(B)\).
  • In the case of Copykwik's photocopy machines, each machine breaking down on a single day is considered independent of the others.
This means that the failure of one machine does not cause or influence another machine to fail. Thus, these events are perfectly qualified as independent events.
Machine Breakdown
In the context of this exercise, a machine breakdown refers to a photocopy machine no longer operating on a given day. For each photocopy machine at Copykwik, there is a given probability that it will cease to function on any particular day.
Each machine has its own specified probability of breakdown:
  • Machine A: \(\frac{1}{50}\)
  • Machine B: \(\frac{1}{60}\)
  • Machine C: \(\frac{1}{75}\)
  • Machine D: \(\frac{1}{40}\)
These probabilities are independent from one another, meaning the likelihood of one machine breaking down does not change because another machine may or may not break down. Understanding these individual probabilities is crucial when trying to determine the probability of an event such as all machines breaking down simultaneously or not breaking down at all.
Event Probability
Event probability refers to the chance or likelihood of a specific event occurring. Each event in probability is assigned a value between 0 and 1.
A probability of 0 indicates an event is impossible, while a probability of 1 means it is certain. You can think of probabilities as fractions, decimals, or percentages to understand how likely an event is to occur.
  • An individual machine breaking down can be considered an event, and its probability has been predefined for each machine based on its performance data.
  • When assessing multiple events, like all machines breaking down, the overall probability is determined by the relation between these individual events.
  • In this exercise, we're interested in the event probability for a combination or absence of events: all four machines breaking down together, and none of them breaking down at all.
This concept helps us predict occurrences or plan for maintenance at Copykwik.
Probability Calculation
Probability calculation involves the mathematical steps used to determine the probability of events. Calculating probability becomes particularly interesting when dealing with multiple events.
To calculate the probability of all independent events occurring together, multiply the individual probabilities:
  • Using the given exercise example, for all machines breaking down, we compute \(P(A \cap B \cap C \cap D) = \frac{1}{50} \times \frac{1}{60} \times \frac{1}{75} \times \frac{1}{40} = \frac{1}{9{,}000{,}000}\).
Similarly, to calculate the probability that none of the machines break down, first determine the probability of each machine not breaking down:
  • For example, \(P(A') = 1 - \frac{1}{50} = \frac{49}{50}\).
Then multiply these results to find the combined probability:\[P(A' \cap B' \cap C' \cap D') = \frac{49}{50} \times \frac{59}{60} \times \frac{74}{75} \times \frac{39}{40} = \frac{167{,}965}{180{,}000}\].
This approach ensures precision and clarity when predicting the likelihood of compound events.

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Most popular questions from this chapter

Based on data obtained from the National Institute of Dental Research, it has been determined that \(42 \%\) of 12 -yr-olds have never had a cavity, \(34 \%\) of 13 -yr-olds have never had a cavity, and \(28 \%\) of 14 -yrolds have never had a cavity. Suppose a child is selected at random from a group of 24 junior high school students that includes six 12 -yr-olds, eight 13 -yr-olds, and ten 14 -yrolds. If this child does not have a cavity, what is the probability that this child is 14 yrs old?

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