/*! 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 22 How many times would you expect ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How many times would you expect to roll a fair die before all 6 sides appeared at least once?

Short Answer

Expert verified
On average, it takes about 6.6 rolls of a fair die to see all six numbers at least once.

Step by step solution

01

Understand the problem and define a discrete random variable

We want to find the average number of rolls it would take for all 6 sides of a fair die to appear at least once. We can define a discrete random variable \(X\) to represent the number of rolls it takes to achieve this.
02

Calculate the probabilities

The probabilities of rolling a specific number on a fair die is constant and equal to \(\dfrac{1}{6}\) for each number, since there are 6 possible outcomes and each outcome is equally likely. However, in our problem, we need a conditional probability - the probability of rolling a specific number given that we haven't rolled all the numbers yet.
03

Use conditional probability

Let's visualize the problem like this: after \(n-1\) rolls, we have seen \(n-1\) distinct numbers, and in the next roll, we want to see the \(n\)-th number for the first time. To calculate this probability, we can use the concept of conditional probability, using the complement rule: \(P(X = n) = P(X \leq n-1) - P(X \leq n-2)\) To calculate \(P(X \leq n-1)\) and \(P(X \leq n-2)\), we can use the fact that rolling a specific number in any roll is independent of the previous rolls, and that the probability of rolling any specific number is constant, equal to \(\dfrac{1}{6}\).
04

Calculate expected value

The expected value \(E(X)\) of a discrete random variable is given by the sum of the product of the values times their probabilities: \(E(X) = \Sigma_{n=1}^{\infty} n \cdot P(X=n)\) In our case, since there are only 6 possible outcomes for each roll, the sum will range from \(n = 6\) to \(n=\infty\). Using the conditional probability formula from Step 3, we can calculate the expected value as follows: \(E(X) = \Sigma_{n=6}^{\infty} n \cdot [P(X \leq n-1) - P(X \leq n-2)]\) \(E(X) = 6 \cdot (1 - 0) + 5 \cdot (\frac{5}{6} - 0) + 4 \cdot (\frac{4}{6} - \frac{3}{6}) + 3 \cdot (\frac{3}{6} - \frac{2}{6}) + 2 \cdot (\frac{2}{6} - \frac{1}{6}) + 1 \cdot (\frac{1}{6})\) \(E(X) \approx 6.6\) So, on average, it takes about 6.6 rolls of a fair die to see all six numbers at least once.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Discrete Random Variable
In probability and statistics, a discrete random variable is a type of random variable that can take on a countable number of distinct values. In this exercise, we use a discrete random variable, which is perfect for modeling scenarios like rolling a die. Here, we define a discrete random variable \(X\) to represent the number of rolls it takes to roll all six sides of a fair die at least once.
Each possible outcome of the dice roll, i.e., each face of the die, corresponds to one of the values that \(X\) can take. This is because each roll results in one of the six integer outcomes. These outcomes are mutually exclusive and include all the possible scenarios, which aligns perfectly with the concept of a discrete random variable.
Conditional Probability
Conditional probability is a fundamental concept in probability theory that helps us understand the likelihood of an event given that another event has already occurred. In this exercise, we utilize conditional probability to determine the likelihood of rolling a specific number on a die, given that not all numbers have appeared at least once.
Imagine you've rolled a fair die several times but still haven't seen one of the numbers show up. The probability of rolling that missing number on the next roll depends on the set of numbers not yet rolled. This is where conditional probability comes in handy. It helps us calculate the likelihood of seeing new numbers as we continue rolling the die.
  • For example, if you've already rolled five unique numbers, the probability of rolling the sixth unique number on the next roll is calculated using the existing results and chances.
  • This leads to better understanding the expected number of rolls needed.
Fair Die
A fair die is a crucial element in understanding rolling probability questions like this one. A fair die means each of its six faces has an equal chance of landing face up. In mathematics, this is depicted as each side having a probability of \(\frac{1}{6}\) of being rolled each time.
A key characteristic of a fair die is that it removes any biases, ensuring that each of the six numbers has an equal opportunity to appear on any given roll. This balance is fundamental in calculating probabilities and expected values accurately, as it ensures no number is more likely to appear than another.
Rolling Probability
Rolling probability is all about understanding the likelihood of rolling anything specific when you toss a die. For a single roll of a fair six-sided die, the probability of rolling any one number is \(\frac{1}{6}\). But things get more interesting when you want to know about sequences of rolls, like when you need all sides to appear.
Rolling probability becomes more complex as we seek the probability of rolling all sides at least once. By incorporating the concept of conditional probability, we observe new patterns and calculate the likelihood of different outcomes based on previously rolled numbers.
  • For instance, the probability that a new side appears decreases slightly with every new and unique number rolled, as there are fewer unknowns left.
  • This can be demonstrated mathematically and helps in finding the expected value of rolls needed.
Understanding these rolling probabilities allows us to grasp the larger picture of randomness and chance when it comes to dice and similar scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent random variables having an unknown continuous distribution function \(F\) and let \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random variables having an unknown continuous distribution function \(G .\) Now order those \(n+m\) variables, and let $$I_{i}=\left\\{\begin{array}{ll}1 & \text { if the } i \text { th smallest of the } n+m \\\& \text { variables is from the } X \text { sample } \\\0 & \text { otherwise }\end{array}\right.$$ The random variable \(R=\sum_{i=1}^{n+m} i I_{i}\) is the sum of the ranks of the \(X\) sample and is the basis of a standard statistical procedure (called the Wilcoxon sum-of-ranks test) for testing whether \(F\) and \(G\) are identical distributions. This test accepts the hypothesis that \(F=G\) when \(R\) is neither too large nor too small. Assuming that the hypothesis of equality is in fact correct, compute the mean and variance of \(R\)

Ten hunters are waiting for ducks to fly by. When a flock of ducks flies overhead, the hunters fire at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability \(.6,\) compute the expected number of ducks that are hit. Assume that the number of ducks in a flock is a Poisson random variable with mean 6

Consider \(n\) independent flips of a coin having probability \(p\) of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if \(n=5\) and the outcome is \(H H T H T,\) then there are 3 changeovers. Find the expected number of changeovers.

Urn 1 contains 5 white and 6 black balls, while urn 2 contains 8 white and 10 black balls. Two balls are randomly selected from urn 1 and are put into urn \(2 .\) If 3 balls are then randomly selected from urn \(2,\) compute the expected number of white balls in the trio. Hint: Let \(X_{i}=1\) if the \(i\) th white ball initially in urn 1 is one of the three selected, and let \(X_{i}=0\) otherwise. Similarly, let \(Y_{i}=1\) if the \(i\) th white ball from urn 2 is one of the three selected, and let \(Y_{i}=0\) otherwise. The number of white balls in the trio can now be written as \(\sum_{1}^{5} X_{i}+\sum_{1}^{8} Y_{i}\)

Consider the following dice game, as played at a certain gambling casino: Players 1 and 2 roll a pair of dice in turn. The bank then rolls the dice to determine the outcome according to the following rule: Player \(i, i=1,2,\) wins if his roll is strictly greater than the bank's. For \(i=\) \(1,2,\) let $$I_{i}=\left\\{\begin{array}{ll}1 & \text { if } i \text { wins } \\\0 & \text { otherwise }\end{array}\right.$$ and show that \(I_{1}\) and \(I_{2}\) are positively correlated. Explain why this result was to be expected.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.