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Chapter 5: Problem 22

Security: Burglar Alarms A large bank vault has several automatic burglar alarms. The probability is \(0.55\) that a single alarm will detect a burglar. (a) Quota Problem How many such alarms should be used for \(99 \%\) certainty that a burglar trying to enter will be detected by at least one alarm? (b) Suppose the bank installs nine alarms. What is the expected number of alarms that will detect a burglar?

Short Answer

Expert verified
(a) 6 alarms; (b) Expected alarms: 4.95.

Step by step solution

01

Define the Problem

We are given that each alarm has a probability of \(0.55\) for detecting a burglar. We need to determine how many alarms are necessary to achieve a \(99\%\) certainty of detection, and the expected number of alarms triggering when nine alarms are installed.
02

Calculate Required Number of Alarms for 99% Detection

To ensure 99% certainty that at least one alarm detects the burglar, we first calculate the probability that none of the alarms detects the burglar: \((1 - 0.55)^n\), where \(n\) is the number of alarms. We set this probability to be less than or equal to 0.01 (for 99% detection), so \ \[(1 - 0.55)^n \leq 0.01.\]
03

Solve for the Required Number of Alarms

We solve the inequality from Step 2. First, compute:\ (0.45)^n \leq 0.01. \ Taking the logarithm of both sides, we have \[ n \cdot \log(0.45) \leq \log(0.01). \] Solve for \(n\): \[ n \geq \frac{\log(0.01)}{\log(0.45)} \approx \frac{-2}{-0.347}\approx 5.75.\] Since \(n\) must be a whole number, round up to \(n=6\) alarms.
04

Calculate Expected Number of Alarms for Nine Installed

With nine alarms installed, each alarm has a 0.55 probability of detecting a burglar. The expected number of alarms that detect a burglar is given by the expression \(E = n \cdot p\), where \(n\) is 9, and \(p\) is 0.55:\ \[E = 9 \times 0.55 = 4.95.\]
05

Conclusion of the Findings

For part (a), to ensure at least 99% certainty of detection, at least 6 alarms are needed. For part (b), if nine alarms are installed, the expected number of alarms that will detect a burglar is 4.95.

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Key Concepts

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

Understanding Expected Value in Probability
Expected value is a fundamental concept in probability and statistics. It's essentially a measure of the center of a probability distribution, which gives us an idea of what to expect in the long run. In simple terms, it's like a weighted average, where outcomes are multiplied by their probabilities, and then these products are summed up. This provides a single value representing the average outcome.
The formula for expected value is:
  • For discrete random variables, the expected value is calculated as: i.e. \( E(X) = \sum x_i \cdot p_i \), where \( x_i \) are the possible values and \( p_i \) are the probabilities of these values.
  • In our burglar alarm example, with nine alarms each having a 0.55 probability of triggering if a burglar is detected, the expected number of alarms that will go off can be found using the formula: \( E = n \cdot p \). That is, \( 9 \times 0.55 = 4.95 \) alarms.
    This means on average, out of nine alarms, about 4.95 alarms will sound.
Solving Exponential Equations
Exponential equations are essential in understanding how certain processes grow or decay over time, and they appear frequently in the field of probability when dealing with compound scenarios. In the context of our burglar alarm problem, solving the number of alarms needed for a 99% certainty involves an exponential equation.
Here's how the process works:
  • The equation for the probability of none of the alarms detecting a burglar is \( (1 - p)^n \), where \( p \) is the probability of a single alarm sounding (0.55 in this case), and \( n \) is the number of alarms.
  • We set this expression to equal 0.01 for 99% certainty: \( (1 - 0.55)^n \leq 0.01 \), indicating the probability of all alarms failing.
  • We solve for \( n \) by employing logarithms: Taking log on both sides gives \( n \cdot \log(0.45) \leq \log(0.01) \).Solving yields \( n \geq 5.75 \), thus rounding up to the nearest integer means \( n = 6 \).
Exponential equations often require taking logarithms for solving, a useful approach when equations involve powers with unknowns.
Navigating Probability Inequalities
Probability inequalities often arises while ensuring certain probabilities do not exceed a specific threshold. They allow us to set conditions and calculate how likely an event is in comparison to a baseline probability. In this problem, probability inequality plays a key role in determining the necessary number of alarms.
Consider the inequality:
  • We start with the probability of no alarms detecting the burglar being \( (1 - 0.55)^n \).
  • To ensure the complement (at least one alarm sounds) is 99% or more, we need: \( (0.45)^n \leq 0.01 \).
  • This inequality ensures that the scenario of not detecting the burglar is highly unlikely.By achieving this through calculation, it confirms that the desired certainty is reached.
Probability inequalities help in assessing risk and ensuring certain levels of safety or certainty, as required. In our example, through the inequality, we ascertain that with 6 alarms, the system meets its security demands.

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Most popular questions from this chapter

Expected Value: Life Insurance Jim is a 60 -year-old Anglo male in reasonably good health. He wants to take out a \(\$ 50,000\) term (that is, straight death benefit) life insurance policy until he is \(65 .\) The policy will expire on his 65 th birthday. The probability of death in a given year is provided by the Vital Statistics Section of the Statistical Abstract of the United States (116th Edition). \begin{tabular}{l|ccccc} \hline\(x=\) age & 60 & 61 & 62 & 63 & 64 \\ \hline\(P\) (death at this age) & \(0.01191\) & \(0.01292\) & \(0.01396\) & \(0.01503\) & \(0.01613\) \\ \hline \end{tabular} Jim is applying to Big Rock Insurance Company for his term insurance policy. (a) What is the probability that Jim will die in his 60 th year? Using this probability and the \(\$ 50,000\) death benefit, what is the expected cost to Big Rock Insurance? (b) Repeat part (a) for years \(61,62,63\), and 64 . What would be the total expected cost to Big Rock Insurance over the years 60 through \(64 ?\) (c) Interpretation If Big Rock Insurance wants to make a profit of \(\$ 700\) above the expected total cost paid out for Jim's death, how much should it charge for the policy? (d) Interpretation If Big Rock Insurance Company charges \(\$ 5000\) for the policy, how much profit does the company expect to make?

Basic Computation: Poisson Distribution Given a binomial experiment with \(n=150\) trials and probability of success on a single trial \(p=0.06\), find the value of \(\lambda\) and then use the Poisson distribution to estimate the probability of \(r \leq 2\) successes.

Statistical Literacy For a binomial experiment, what probability distribution is used to find the probability that the first success will occur on a specified trial?

Basic Computation: Binomial Distribution Consider a binomial experiment with \(n=7\) trials where the probability of success on a single trial is \(p=0.60\). (a) Find \(P(r=7)\). (b) Find \(P(r \leq 6)\) by using the complement rule.

Combination of Random Variables: Golf Norb and Gary are entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations. $$ \text { Norb, } x_{1}: \mu_{1}=115 ; \sigma_{1}=12 \quad \text { Gary, } x_{2}: \mu_{2}=100 ; \sigma_{2}=8 $$ In the tournament, Norb and Gary are not playing together, and we will assume their scores vary independently of each other. (a) The difference between their scores is \(W=x_{1}-x_{2} .\) Compute the mean, variance, and standard deviation for the random variable \(W\). (b) The average of their scores is \(W=0.5 x_{1}+0.5 x_{2}\). Compute the mean, variance, and standard deviation for the random variable \(W\). (c) The tournament rules have a special handicap system for each player. For Norb, the handicap formula is \(L=0.8 x_{1}-2 .\) Compute the mean, variance, and standard deviation for the random variable \(L .\) (d) For Gary, the handicap formula is \(L=0.95 x_{2}-5 .\) Compute the mean, variance, and standard deviation for the random variable \(L .\)
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