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Approximately \(30 \%\) of the calls to an airline reservation phone line result in a reservation being made. a. Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls result in a reservation? b. What assumption did you make to calculate the probability in Part (a)? c. What is the probability that at least one call results in a reservation being made?

Short Answer

Expert verified
The probability that no calls result in a reservation, the assumptions made, and the probability at least one call results in a reservation are calculated as described in the steps above. This question is an example of binomial distribution concepts in statistics.

Step by step solution

01

Identifying the variables

Firstly, we need to identify the variables in the problem: Number of trials (\( n \)) is 10 (ten phone calls), Success probability (\( p \)) is 0.30 (the probability that a call results in a reservation), and we're asked to find the probability of no successful calls (Reservation made, \( k \) is 0). Siunce we're dealing with a binomial distribution, the formula to be used is: \[ P(x=k) = C(n, k) * p^k * (1-p)^{n-k}\]
02

Calculate the probability of no successful calls

Plug in the given values into the formula: \( C(10, 0) * 0.30^0 * (1-0.30)^{10-0} \). The term \( C(10, 0) \) represents the combinations of 10 (phone calls) taken 0 at a time, which is 1. And any number to the power of 0 is 1. Hence the equation simplifies to \( (1-0.30)^{10} \). Solving it will give us the required probability for Part (a).
03

Assumption made for the calculation

The assumption that is made here refers to the fundamental principles of a binomial distribution, which are: 1) The trials are independent; the result of one does not affect the result of another. 2) The probability of success is the same for each trial. That's the answer for Part (b)
04

Calculate the probability of at least one successful call

The probability that at least one call results in a reservation can be calculated as the total probability (which is 1) minus the probability of no reservations (no successful calls). So, use the value calculated in Step 2, subtract it from 1, and this will give the final answer for Part (c).

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics that deals with quantifying the likelihood of events. It is used in various fields to predict outcomes and make decisions under uncertainty. Understanding probability is crucial when dealing with situations that have several possible results, especially when those outcomes are random. In the case of the airline reservation phone line, probability theory helps us calculate the chances of an operator making a reservation during calls.

In probability theory, the sum of the probabilities of all possible outcomes of a random trial must equal 1. This is because one of those outcomes must occur. For example, if we define the success of an operator as making a reservation (which happens 30% of the time), there is always a 70% chance that a call will not result in a reservation. Therefore, the probabilities of these two outcomes (success or no success) should add up to 100%.
Independent Trials
Independent trials are a fundamental concept in probability, especially in the context of the binomial probability distribution. It refers to the scenario where the outcome of one trial does not affect the outcomes of subsequent trials. For each call the operator handles, it is assumed that whether or not a reservation is made does not depend on the results of the prior calls. Each call is treated as a separate event.

This concept is critical in calculating the probabilities of different outcomes since it allows for the use of certain probabilistic models, like the binomial distribution. The assumption of independence underlies many statistical procedures and is a part of the statistical assumptions that were made for the calculations in the given exercise.
Statistical Assumptions
Statistical assumptions are the set of conditions under which certain statistical models can be applied. In binomial probability distribution problems such as the reservation call scenario, we make two key assumptions. First, each trial is independent, which means no call's outcome influences another. Consequently, this supports the use of the binomial formula for our calculations. Second, the probability of success is consistent across all calls, described as a 'success probability'.

With these assumptions, we can use mathematical models to predict probabilities accurately. However, if these assumptions do not hold—if the trials are not truly independent or the probability of success varies—it may lead to incorrect predictions and a need for a different model or a modified approach.
Success Probability
Success probability, denoted as 'p' in the binomial distribution formula, is the likelihood of a single trial resulting in a success. For our scenario, a success is defined as an operator making a reservation, which is given as 30%. This figure is essential in computing the overall probability of a certain number of successes over a series of trials. In binomial distribution, this probability must remain constant throughout all trials, which is one of the model's fundamental assumptions.

The complement of the success probability is the probability of failure (1-p), which also features in calculations. In our exercise, the probability of not making a reservation in a single call is 70% (1 - 0.30). This concept helps us understand the dynamics of the likely outcomes from a sequence of independent trials and calculate the probability of different numbers of successes across those trials.

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Most popular questions from this chapter

Is ultrasound a reliable method for determining the gender of an unborn baby? The accompanying data on 1000 births are consistent with summary values that appeared in the online version of the Journal of Statistics Education ("New Approaches to Leaming Probability in the First Statistics Course" [2001]). $$ \begin{array}{ccc} & \begin{array}{c} \text { Ultrasound } \\ \text { Predicted } \\ \text { Female } \end{array} & \begin{array}{c} \text { Ultrasound } \\ \text { Predicted } \\ \text { Male } \end{array} \\ \hline \begin{array}{c} \text { Actual Gender Is } \\ \text { Female } \end{array} & 432 & 48 \\ \begin{array}{c} \text { Actual Gender Is } \\ \text { Male } \end{array} & 130 & 390 \\ \hline \end{array} $$ a. Use the given information to estimate the probability that a newborn baby is female, given that the ultrasound predicted the baby would be female. b. Use the given information to estimate the probability that a newborn baby is male, given that the ultrasound predicted the baby would be male. c. Based on your answers to Parts (a) and (b), do you think that a prediction that a baby is male and a prediction that a baby is female are equally reliable? Explain.

A single-elimination tournament with four players is to be held. In Game 1, the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3, the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: \(P(\) seed 1 defeats seed 4\()=.8\) \(P(\) seed 1 defeats seed 2\()=.6\) \(P(\) seed 1 defeats seed 3\()=.7\) \(P(\) seed 2 defeats seed 3\()=.6\) \(P(\) seed 2 defeats seed 4\()=.7\) \(P(\operatorname{seed} 3\) defeats seed 4\()=.6\) a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate Game 3 in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\).) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.

An article in the New york Times (March 2. 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\), and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F\) ? b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F)\) ? A Venn diagram might help.)

Three friends \((A, B\), and \(C)\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats C) \(=.8, P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that \(\mathrm{A}\) wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? b. What is the probability that \(A\) wins both her matches? c. What is the probability that \(A\) loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

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