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Chapter 5: Problem 16

The probability that a single radar station will detect an enemy plane is \(0.65\). (a) Quota Problem How many such stations are required to be \(98 \%\) certain that an enemy plane flying over will be detected by at least one station? (b) If four stations are in use, what is the expected number of stations that will detect an enemy plane?

Short Answer

Expert verified
(a) 6 stations are required to be 98% certain of detection. (b) The expected number of stations that will detect the plane is 2.6.

Step by step solution

01

Understanding the Detection Probability

The probability that a single radar station detects an enemy plane is given as 0.65. This means that the probability that a radar station does not detect the enemy plane is \( 1 - 0.65 = 0.35 \). This concept will be important for solving part (a) of the problem.
02

Step for Part (a): Using Complementary Probability

To find how many stations are needed to be 98% certain that at least one will detect the plane, we'll use the complementary probability principle. Let's call the number of stations needed \( n \). The probability that not a single station detects the plane is \( 0.35^n \). We want the probability of the complement (at least one detection) to be 0.98, hence,\[ 1 - 0.35^n = 0.98 \] which simplifies to \[ 0.35^n = 0.02 \].
03

Step for Part (a): Solving the Equation

To solve the equation \( 0.35^n = 0.02 \), we'll take the natural logarithm of both sides,\[ \ln(0.35^n) = \ln(0.02) \]Applying the power rule for logarithms, this becomes\[ n \cdot \ln(0.35) = \ln(0.02) \].Thus, solving for \( n \), we have \[ n = \frac{\ln(0.02)}{\ln(0.35)} \].Calculating this gives \( n \approx 5.32 \). Since the number of stations must be a whole number, we round up to 6.
04

Step for Part (b): Calculating Expected Value

Given 4 radar stations, each having a probability of 0.65 of detecting the plane, we calculate the expected number of stations that detect the plane (denoted as \( E(X) \)). For each station, the expected number of detections is 0.65. Thus, \[ E(X) = 4 \times 0.65 = 2.6 \].

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Key Concepts

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

Detection Probability
When we talk about detection probability, we are referring to the likelihood that a certain event, such as the detection of an enemy plane by a radar station, will occur. In our scenario, each radar station has a detection probability of 0.65. This means that there's a 65% chance for a radar station to successfully sense the presence of an enemy plane flying over.

Understanding detection probability is crucial for making strategic decisions, like determining how many radars are necessary to achieve a specific confidence level in plane detection. If each radar independently functions with a 65% success rate, the collective probability increases when multiple devices are utilized, thereby improving overall performance.

Here's a quick recap:
  • Detection Probability = Probability that a radar station detects a plane.
  • In this case: 0.65 or 65%.
  • The probability that a radar fails to detect the plane is 1 minus the detection probability, which is 0.35 or 35%.
Expected Value
The expected value is a fundamental concept in probability theory that provides the average outcome of a random variable across numerous trials. It predicts the mean result you can anticipate after many experiments.

For part (b) of the exercise, we calculate how many of the four radar stations are expected to detect an enemy plane. Since each station operates with a probability of 0.65, we use the expected value formula for a binomial distribution: \( E(X) = n \cdot p \). Here, \( n = 4 \) (number of stations) and \( p = 0.65 \) (probability of detection).

Thus, the expected value is:
  • \( E(X) = 4 \times 0.65 = 2.6 \)
This calculation implies that on average, 2.6 out of 4 stations will detect the plane. Although you can't have a part of a station detecting a plane, this number provides a statistical expectation over many similar trials.

By understanding expected value, we quantify outcomes and can estimate performance levels when facing uncertainty. It's a predictive tool used extensively in decision making.
Complementary Probability
The concept of complementary probability is about finding the probability of the opposite event. If you are not sure about the occurrence of an event but know the probability of it not happening, then complementary probability helps you find the original probability.

Using part (a) of the exercise, we rely on this concept to determine how many radar stations are needed for a 98% chance that at least one of them will detect the airplane. If the probability that one station fails is 0.35, the probability that all \( n \) stations will fail simultaneously is \( 0.35^n \).

To be 98% certain that at least one station will detect the plane, we set this complementary probability to 0.02:
  • \( 1 - 0.35^n = 0.98 \)
  • \( 0.35^n = 0.02 \)
Then, solving for \( n \), we use logarithms to find that approximately 6 stations are needed. Complementary probability is useful for problems where it's easier to calculate the probability of an event not happening, and it significantly aids in decision-making under risk conditions.

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Most popular questions from this chapter

Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about \(57 \%\) of all people who take the state bar exam pass (Source: The Book of \(\mathrm{Odds}\) by Shook and Shook, Signet). Let \(n=1,2,3, \ldots\) represent the number of times a person takes the bar exam until the first pass. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Bob first passes the bar exam on the second try \((n=2) ?\) (c) What is the probability that Bob needs three attempts to pass the bar exam? (d) What is the probability that Bob needs more than three attempts to pass the bar exam? (e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

A research team at Cornell University conducted a study showing that approximately \(10 \%\) of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions (Source: Chances: Risk and Odds in Everyday Life, by James Burke). At a board meeting of 20 businessmen, all of whom wear ties, what is the probability that (a) at least one tie is too tight? (b) more than two ties are too tight? (c) no tie is too tight? (d) at least 18 ties are not too tight?

In the western United States, there are many dry land wheat farms that depend on winter snow and spring rain to produce good crops. About \(65 \%\) of the years there is enough moisture to produce a good wheat crop, depending on the region (Reference: Agricultural Statistics, United States Department of Agriculture). (a) Let \(r\) be a random variable that represents the number of good wheat crops in \(n=8\) years. Suppose the Zimmer farm has reason to believe that at least 4 out of 8 years will be good. However, they need at least 6 good years out of 8 years to survive financially. Compute the probability that the Zimmers will get at least 6 good years out of 8, given what they believe is true; that is, compute \(P(6 \leq r \mid 4 \leq r) .\) See part \((\mathrm{d})\) for a hint. (b) Let \(r\) be a random variable that represents the number of good wheat crops in \(n=10\) years. Suppose the Montoya farm has reason to believe that at least 6 out of 10 years will be good. However, they need at least 8 good years out of 10 years to survive financially. Compute the probability that the Montoyas will get at least 8 good years out of 10, given what they believe is true; that is, compute \(P(8 \leq r \mid 6 \leq r)\). (c) List at least three other areas besides agriculture to which you think conditional binomial probabilities can be applied. (d) Hint for solution: Review item 6, conditional probability, in the summary of basic probability rules at the end of Section \(4.2 .\) Note that $$ P(A \mid B)=\frac{P(A \text { and } B)}{P(B)} $$ and show that in part (a) $$ P(6 \leq r \mid 4 \leq r)=\frac{P((6 \leq r) \text { and }(4 \leq r))}{P(4 \leq r)}=\frac{P(6 \leq r)}{P(4 \leq r)} $$

In an experiment, there are \(n\) independent trials. For each trial, there are three outcomes, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). For each trial, the probability of outcome \(\mathrm{A}\) is \(0.40\); the probability of outcome \(\mathrm{B}\) is \(0.50\); and the probability of outcome \(\mathrm{C}\) is \(0.10\). Suppose there are 10 trials. (a) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type \(\mathrm{B}\), and one of type C? Explain. (b) Can we use the binomial experiment model to determine the probability of four outcomes of type \(\mathrm{A}\) and six outcomes that are not of type A? Explain. What is the probability of success on each trial?

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing \(77 \%\) of the students in Western Civilization each term. Let \(n=1,2,3, \ldots\) represent the number of times a student takes Western Civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Susan passes on the first try \((n=1) ?\) (c) What is the probability that Susan first passes on the second try \((n=2) ?\) (d) What is the probability that Susan needs three or more tries to pass Western Civilization? (e) What is the expected number of attempts at Western Civilization Susan must make to have her (first) pass? Hint Use \(\mu\) for the geometric distribution and round.
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