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Consider two binomial experiments. a. The first binomial experiment consists of six trials. How many outcomes have exactly one success, and what are these outcomes? b. The second binomial experiment consists of 20 trials. How many outcomes have exactly 10 successes? exactly 15 successes? exactly 5 successes?

Short Answer

Expert verified
For the first experiment, there are 6 outcomes with exactly one success. In the second experiment, there are 184756 outcomes with exactly 10 successes, 15504 outcomes with exactly 15 successes, and 15504 outcomes with exactly 5 successes.

Step by step solution

01

Determine outcomes for the first experiment

For the first experiment, there are six trials and the problem asks to find outcomes having exactly one success. Use the formula for combination \[ C(n,k) = \frac{n!}{k!(n-k)!} \]. When n is 6 and k is 1, it becomes \( C(6,1) = 6 \), which means there are 6 outcomes that represent exactly one success.
02

Determine outcomes for the second experiment

For the second experiment, again use the combination formula to calculate outcomes with exactly 10, 15 and 5 successes, while there are total 20 trials.For 10 successes, it's \( C(20,10) = 184756 \).For 15 successes, it's \( C(20,15) = 15504 \).And finally, for 5 successes, it's \( C(20,5) = 15504 \). Each calculation uses the formula \[ C(n,k) = \frac{n!}{k!(n-k)!} \] with n as 20 and varying values of k.

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Key Concepts

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

Combinations
Combinations are an essential concept when dealing with problems where the arrangement order does not matter. This is particularly important in binomial experiments, where you might be interested in finding how many ways you can choose a subset of items from a larger set. For example, in a binomial experiment with multiple trials, we calculate the number of ways of obtaining a certain number of successes in a fixed number of trials.

To compute combinations, we use the formula for combination:
  • \(C(n,k) = \frac{n!}{k!(n-k)!}\)
Here, \( n \) is the total number of items, \( k \) is the number of items to choose, and \(!\) represents a factorial. The factorial of a number \( n \) is the product of all positive integers up to \( n \).

This formula can be applied to various situations in probability, aiding in calculating outcomes with specific properties, such as exactly one success, which was demonstrated in the original exercise.
Binomial Coefficient
The binomial coefficient is directly related to combinations and is often visualized in the context of binomial expansions. In practice, the binomial coefficient tells us the number of ways to choose \( k \) successes out of \( n \) trials in a binomial experiment. It is denoted by \( C(n,k) \), and its calculation involves using the same formula we use for combinations.
Let's further illustrate this: given an experiment with 20 trials, to find how many outcomes have exactly 10 successes, you compute \(C(20,10) = \frac{20!}{10! \cdot (20-10)!} = 184756\).
The binomial coefficient supports various areas of combinatorics beyond just calculating probabilities in binomial experiments, making it a fundamental element of discrete mathematics.
Probability Theory
Probability Theory is the mathematical framework that allows us to analyze and predict random events, a cornerstone for understanding binomial experiments. It involves calculating the likelihood of various outcomes, providing a structured approach to uncertainty. In binomial experiments, each trial is independent and has only two possible outcomes: success or failure.
In these experiments, probabilities are calculated using both the concept of combinations and the binomial distribution formula, which considers the success probability \( p \) and the number of trials \( n \). This formula helps in determining the probability of a specific number of successes out of \( n \) trials:
  • \(P(X=k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}\)
This equation captures the beauty of probability theory by quantifying how likely it is to achieve \( k \) successes, considering the inherent randomness of the process.
Understanding these concepts is crucial not only for solving textbook problems but also for comprehending real-world random processes.

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Most popular questions from this chapter

An appliance dealer sells three different models of upright freezers having \(13.5,15.9\), and \(19.1\) cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & .2 & .5 & .3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). b. If the price of the freezer depends on the size of the storage space, \(x\), such that Price \(=25 x-8.5\), what is the mean value of the variable Price paid by the next customer? c. What is the standard deviation of the price paid?

A coin is spun 25 times. Let \(x\) be the number of spins that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. What is the probability of judging the coin fair when \(P(\mathrm{H})=.9\), so that there is a substantial bias? Repeat for \(P(\mathrm{H})=.1 .\) c. What is the probability of judging the coin fair when \(P(\mathrm{H})=.6\) ? when \(P(\mathrm{H})=.4 ?\) Why are the probabilities so large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and (b) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

The probability distribution of \(x\), the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lrrrrr} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & .54 & .16 & .06 & .04 & .20 \end{array} $$ a. Calculate the mean value of \(x\). b. What is the probability that \(x\) exceeds its mean value?

Because \(P(z<.44)=.67,67 \%\) of all \(z\) values are less than \(.44\), and \(.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91 st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the 30 th \(z\) percentile?

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the variable \(x\) as the number of people who actually show up for a sold-out flight. From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{lrrrrrrrr} x & 95 & 96 & 97 & 98 & 99 & 100 & 101 & 102 \\ p(x) & .05 & .10 & .12 & .14 & .24 & .17 & .06 & .04 \\ x & 103 & 104 & 105 & 106 & 107 & 108 & 109 & 110 \\ p(x) & .03 & .02 & .01 & .005 & .005 & .005 & .0037 & .0013 \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

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