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Chapter 5: Problem 20

The probability that a single radar station will detect an enemy plane is 0.65 (a) How many such stations are required for \(98 \%\) certainty 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) At least 4 stations are required for 98% certainty. (b) Approximately 2.6 stations will detect the plane.

Step by step solution

01

Understanding the Problem

We have two parts in this exercise. In part (a), we need to calculate the minimum number of radar stations required so that the probability of detecting an enemy plane by at least one station is at least 98%. In part (b), we need to find the expected number of stations out of four that will detect an enemy plane.
02

Calculating for Part (a)

To find the number of stations required (n) for a 98% certainty, we can use the complement probability: the plane goes undetected. The probability that a single radar does not detect the plane is 1 - 0.65 = 0.35. The probability that n stations do not detect the plane is 0.35^n. We want the complement of this probability to be 0.98:\[ 1 - 0.35^n = 0.98 \]Solving for n:\[ 0.35^n = 0.02 \]Taking the logarithm:\[ \log(0.35^n) = \log(0.02) \]\[ n \cdot \log(0.35) = \log(0.02) \]\[ n = \frac{\log(0.02)}{\log(0.35)} \]By calculating:\[ n \approx \frac{-1.69897}{-0.45593} \approx 3.72 \]Therefore, we need at least 4 stations since we can't have a fraction of a station.
03

Calculating for Part (b)

For the second part, we know we have 4 stations, and each station independently detects a plane with a probability of 0.65. The expected number of stations detecting the plane is given by the total number of trials times the probability of success in each trial:\[ E = n \cdot p = 4 \cdot 0.65 = 2.6 \]Thus, we expect about 2.6 stations, out of the 4, to detect the plane.

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Key Concepts

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

Complement Probability
In probability theory, the complement probability is an important concept that assists in determining the probability of at least one event occurring.
It is particularly useful when calculating the probability that an event does not happen, and then subtracting that probability from 1 to find the probability that the event does occur.
For example, in our exercise, we began by determining the probability that a single radar station does not detect an enemy plane. This value is equal to 1 minus the probability that it does detect the plane, which is given as 0.65.
So, the complement probability is calculated as follows:
  • The probability that a radar station misses a plane is: \[ 1 - 0.65 = 0.35 \]
To find the probability that multiple stations do not detect the plane, we raise the complement probability to the power of the number of stations. With this method, we calculate the smallest number of stations necessary for a 98% detection probability, by ensuring that the complement probability (the probability of missing detection) is no more than 2%.
Using the formula:
  • \[ 1 - 0.35^n = 0.98 \]
We solve for \( n \) to ensure we meet the requirement of 98% certainty.
Expected Value
Expected value in probability gives an average outcome if an experiment or situation is repeated over and again.
It's a powerful concept as it provides insights into what we should anticipate on average, although actual outcomes on any given trial might vary.
In our exercise, the question was to find out how many radar stations, out of four, are expected to detect an enemy plane. Each station independently detects the plane with a probability of 0.65.
To compute the expected number, you multiply the total number of trials (stations, in this case) by the probability of success (detecting the plane) in one trial.
  • The expected value is calculated as:\[E = n \cdot p = 4 \cdot 0.65 = 2.6\]
This result tells us that we can expect approximately 2.6 out of 4 stations to successfully detect the enemy plane on average.
Independent Events
Events are considered independent when the outcome of one event does not affect the outcome of another.
Understanding this is crucial when dealing with probability scenarios involving multiple agents or trials, like multiple radar stations.
In our context, each radar station operates independently of the others, meaning the ability of one station to detect a plane does not influence the ability of others in any way.
  • This independence allows us to directly multiply probabilities:
For example, when determining the expected number of successful detections using four independent radar stations, the probability calculation for one is unaffected by others.
Independent operations simplify these kinds of problems, allowing us to use multiplicative rules to determine the combined effect, as seen in both calculating complement probabilities and expected values.
Thus, this understanding is pivotal in accurately analyzing scenarios with multiple independent agents or trials.

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Most popular questions from this chapter

In his doctoral thesis, L. A. Beckel (University of Minnesota, 1982 ) studied the social behavior of river otters during the mating season. An important role in the bonding process of river otters is very short periods of social grooming. After extensive observations, Dr. Beckel found that one group of river otters under study had a frequency of initiating grooming of approximately 1.7 for every 10 minutes. Suppose that you are observing river otters for 30 minutes. Let \(r=0,1,2, \ldots\) be a random variable that represents the number of times (in a 30-minute interval) one otter initiates social grooming of another. (a) Explain why the Poisson distribution would be a good choice for the probability distribution of \(r\) What is \(\lambda\) ? Write out the formula for the probability distribution of the random variable \(r\) (b) Find the probabilities that in your 30 minutes of observation, one otter will initiate social grooming four times, five times, and six times. (c) Find the probability that one otter will initiate social grooming four or more times during the 30 -minute observation period. (d) Find the probability that one otter will initiate social grooming fewer than four times during the 30 -minute observation period.

For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

Aircraft inspectors (who specialize in mechanical engineering) report wing cracks in aircraft as nonexistent, detectable (but still functional), or critical (needs immediate repair). For a particular model of commercial jet 10 years old, history indicates \(75 \%\) of the planes had no wing cracks, \(20 \%\) had detectable wing cracks, and \(5 \%\) had critical wing cracks. Five planes that are 10 years old are randomly selected. What is the probability that (a) 4 have no cracks, 1 has detectable cracks, and 0 have no critical cracks? (b) 3 have no cracks, I has detectable cracks, and I has critical cracks?

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 \(L.\)

Given a binomial experiment with probability of success on a single trial \(p=0.40,\) find the probability that the first success occurs on trial number \(n=3\)
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