/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A die is rolled 30 times. What i... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 8

A die is rolled 30 times. What is the probability that a 6 turns up exactly 5 times? What is the most probable number of times that a 6 will turn up?

Short Answer

Expert verified
The probability of rolling a six exactly 5 times is approximately 0.139. The most likely number of sixes is 5.

Step by step solution

01

Define the Variables

Consider the problem as a Binomial distribution since we roll a die 30 times and are interested in the occurrence of a 6. Let \( n = 30 \) (number of trials), \( k = 5 \) (number of successful outcomes we are interested in), and \( p = \frac{1}{6} \) (probability of rolling a six).
02

Use the Binomial Probability Formula

To find the probability of getting exactly 5 sixes in 30 rolls, we use the formula for Binomial probability: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Plug in the given values:\[ P(X = 5) = \binom{30}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^{25} \]
03

Calculate the Binomial Coefficient

Compute the binomial coefficient \( \binom{30}{5} \), which represents combinations of 30 items taken 5 at a time:\[ \binom{30}{5} = \frac{30!}{5! (30-5)!} = \frac{30 \cdot 29 \cdot 28 \cdot 27 \cdot 26}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 142506 \]
04

Calculate the Probability Components

Calculate the probabilities of success and failure:\[ \left( \frac{1}{6} \right)^5 = \frac{1}{7776} \] and\[ \left( \frac{5}{6} \right)^{25} \] To approximate \( \left( \frac{5}{6} \right)^{25} \), we can use a calculator for precise values.
05

Compute the Final Probability

Now compute the full binomial probability as: \[ P(X = 5) = 142506 \times \frac{1}{7776} \times \left( \frac{5}{6} \right)^{25} \] Using a calculator, find this value to be approximately 0.139.
06

Find the Most Likely Number of Sixes

The most probable number of sixes is given by the expected value \( E = np \), where \( n = 30 \) and \( p = \frac{1}{6} \). Compute \( E = 30 \times \frac{1}{6} = 5 \). This suggests that the most likely number of sixes in 30 rolls is 5.

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Key Concepts

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

Probability
In the context of a binomial distribution, probability is an essential concept. It tells us how likely it is for a particular event to happen. When dealing with a six-sided die, the probability of rolling a six in one toss is \( \frac{1}{6} \). This means that there is a one in six chance that a six will appear.

To find the probability of rolling exactly five sixes in 30 rolls, we use the binomial probability formula. This considers the number of ways the event (rolling a six) can occur and the likelihood of each occurrence. The probability formula is:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
  • Where \( n \) is the total number of trials (30), \( k \) is the number of successful outcomes (5), and \( p \) is the probability of success in each trial (\( \frac{1}{6} \)).
Understanding how to apply this formula helps in calculating the exact probability of an event happening under binomial conditions.
Binomial Coefficient
The binomial coefficient is a crucial part of the binomial probability formula. It describes the number of ways to choose \( k \) successes from \( n \) trials. Mathematically, it is represented as \( \binom{n}{k} \), which is read as "n choose k." In the context of rolling a die 30 times and getting exactly 5 sixes, the binomial coefficient is \( \binom{30}{5} \).

Calculating this involves factorials, which are the products of all positive integers up to a certain number. The formula for the binomial coefficient is:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
  • For the given example, \( \binom{30}{5} = \frac{30!}{5! \times 25!} \)
  • The result, 142506, indicates how many possible ways there are to achieve exactly 5 sixes in 30 rolls of a die.
This calculation is key to determining the overall probability in a binomial experiment.
Expected Value
The expected value in a binomial distribution gives a measure of the center or average outcome expected from an experiment across many trials. It is the product of the number of trials \( n \) and the probability of success \( p \) in each trial. For the die rolling scenario, the expected value indicates the most likely number of sixes across 30 rolls.

The formula for expected value is:
  • \( E = np \)
  • In the exercise, \( n = 30 \) and \( p = \frac{1}{6} \), so \( E = 30 \times \frac{1}{6} = 5 \)
  • This means that, on average, you can expect to roll a six 5 times in 30 tries.
Understanding expected value helps to see the bigger picture of what results are typical, highlighting the most probable outcomes in random processes. It is a foundational concept in probability and statistics that simplifies the analysis of complex probabilistic scenarios.

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Most popular questions from this chapter

Let \(n(\bmod m)\) denote the remainder when the integer \(n\) is divided by the integer \(m\). Write a computer program to compute the numbers \(\left(\begin{array}{c}n \\ j\end{array}\right)(\bmod m)\) where \(\left(\begin{array}{l}n \\ j\end{array}\right)\) is a binomial coefficient and \(m\) is an integer. You can do this by using the recursion relations for generating binomial coefficients, doing all the arithmetic using the basic function \(\bmod (n, m)\). Try to write your program to make as large a table as possible. Run your program for the cases \(m=2\) to 7 . Do you see any patterns? In particular, for the case \(m=2\) and \(n\) a power of 2 , verify that all the entries in the \((n-1)\) st row are 1 . (The corresponding binomial numbers are odd.) Use your pictures to explain why this is true.

A baseball player, Smith, has a batting average of .300 and in a typical game comes to bat three times. Assume that Smith's hits in a game can be considered to be a Bernoulli trials process with probability .3 for success. Find the probability that Smith gets \(0,1,2,\) and 3 hits.

There are three different routes connecting city A to city B. How many ways can a round trip be made from \(A\) to \(B\) and back? How many ways if it is desired to take a different route on the way back?

There are \(n\) applicants for the director of computing. The applicants are interviewed independently by each member of the three-person search committee and ranked from 1 to \(n\). A candidate will be hired if he or she is ranked first by at least two of the three interviewers. Find the probability that a candidate will be accepted if the members of the committee really have no ability at all to judge the candidates and just rank the candidates randomly. In particular, compare this probability for the case of three candidates and the case of ten candidates.

You are playing heads or tails with Prosser but you suspect that his coin is unfair. Von Neumann suggested that you proceed as follows: Toss Prosser's coin twice. If the outcome is HT call the result win. if it is TH call the result lose. If it is TT or HH ignore the outcome and toss Prosser's coin twice again. Keep going until you get either an HT or a TH and call the result win or lose in a single play. Repeat this procedure for each play. Assume that Prosser's coin turns up heads with probability \(p\). (a) Find the probability of \(\mathrm{HT}, \mathrm{TH}, \mathrm{HH}, \mathrm{TT}\) with two tosses of Prosser's coin. (b) Using part (a), show that the probability of a win on any one play is \(1 / 2\), no matter what \(p\) is.
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