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Three people work independently at deciphering a message in code. The probabilities that they will decipher it are \(\frac{1}{5}, \frac{1}{4}\) and \(\frac{1}{3}\). What is the probability that the message will be deciphered?

Short Answer

Expert verified
The probability that the message will be deciphered is \( \frac{3}{5} \).

Step by step solution

01

Understanding the Problem

We need to find the probability that at least one of the three individuals will succeed in deciphering the message. Since they work independently, each has a distinct probability of success: \( \frac{1}{5} \), \( \frac{1}{4} \), and \( \frac{1}{3} \).
02

Calculate the Probability of Each Person Failing

The probability of each person failing to decipher the message is the complement of their success probability. Thus, the probabilities are: \( 1 - \frac{1}{5} = \frac{4}{5} \), \( 1 - \frac{1}{4} = \frac{3}{4} \), and \( 1 - \frac{1}{3} = \frac{2}{3} \).
03

Calculate the Probability That All Fail

Since they work independently, the probability that all three fail is the product of their individual failure probabilities: \( \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{4}{5} \times \frac{2}{3} = \frac{2}{5} \).
04

Calculate the Probability That at Least One Succeeds

The probability that at least one of them will succeed is the complement of all failing: \( 1 - \frac{2}{5} = \frac{3}{5} \).
05

Final Answer

Thus, the probability that the message will be deciphered is \( \frac{3}{5} \).

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Key Concepts

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

Complementary Probability
When dealing with probabilities, complementary probability is an essential concept to understand. It refers to the probability of the opposite event occurring. Simply put, if you know the probability of an event, the complementary probability tells you how likely it is that the event won’t happen. The sum of the probability of an event and its complement is always 1. This is because something will either happen or it won't—there's no middle ground.
For example, if the probability of a person deciphering a code is \( \frac{1}{5} \), the probability of them not deciphering it is \( 1 - \frac{1}{5} = \frac{4}{5} \). Complementary probabilities are particularly useful when calculating complex events, as they allow you to focus on the scenarios where the event does not occur.
Understanding and using complementary probabilities can simplify many probability questions. Instead of figuring out every detail, you can determine the likelihood of the opposite scenario and subtract it from 1.
Independent Events
In probability theory, independent events are those events whose occurrence or outcome does not affect the probability of the other events. This means, if Event A occurs, it has no bearing on whether Event B occurs, and vice versa.
Independence is a common assumption that simplifies the calculation of probabilities. In the exercise, the probability of each person deciphering the message is independent of the others. Each person has their own separate probability of success \( \frac{1}{5}, \frac{1}{4}, \) and \( \frac{1}{3} \), respectively.
Understanding whether events are independent is crucial because it informs how probabilities are calculated. If events are dependent, then the occurrence of one affects the occurrence of the other, and different rules apply.
In cases where events are independent, calculating probabilities can often involve multiplication, as each event's probability remains constant regardless of the others.
Probability Multiplication Rule
The probability multiplication rule is used to find the joint probability of multiple independent events occurring. For independent events, the rule states that the probability of all events occurring is the product of their individual probabilities. This is because each event occurs without influencing the others.
For example, in the exercise, the goal is to find the probability that all individuals fail to decipher the message. We calculate this by multiplying the probabilities of each person failing: \( \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \). Once calculated, this gives the overall probability of all three failing together.
This rule is particularly helpful for complex problems with multiple events, as it simplifies the process of finding the probability of concurrent independent events.
  • Always ensure the events are truly independent before applying the rule.
  • The rule can dramatically streamline solving probability problems by breaking them into manageable parts.

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Most popular questions from this chapter

A manufacturer has agreed to dispatch small servomechanisms in cartons of 100 to a distributor. The distributor requires that \(90 \%\) of cartons contain at most one defective servomechanism. Assuming the Poisson approximation to the binomial distribution, write down an equation for the Poisson parameter \(\lambda\) such that the distributor's requirements are just satisfied. Solve by trial and error (approximate solution \(0.5\) ), and hence find the required proportion of manufactured servomechanisms that must be satisfactory.

The City Engineer's department installs 10000 fluorescent lamp bulbs in street lamp standards. The bulbs have an average life of 7000 operating hours with a standard deviation of 400 hours. Assuming that the life of the bulbs, \(L\), is a normal random variable, what number of bulbs might be expected to have failed after 6000 operating hours? If the engineer wishes to adopt a routine replacement policy which ensures that no more than \(5 \%\) of the bulbs fail before their routine replacement, after how long should the bulbs be replaced?

The probability of issuing a drill of high brittleness (a reject) is \(0.02\). Drills are packed in boxes of 100 each. What is the probability that the number of defective drills is no greater than two?

The diameter of ball-bearings produced by a machine is a random variable having a normal distribution with mean \(6.00 \mathrm{~mm}\) and standard deviation \(0.02 \mathrm{~mm}\). If the diameter tolerance is \(\pm 1 \%\), find the proportion of ball-bearings produced that are out of tolerance. After several years' use, machine wear has the effect of increasing the standard deviation, although the mean diameter remains constant. The manufacturer decides to replace the machine when \(2 \%\) of its output is out of tolerance. What is the standard deviation when this happens?

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