/*! 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 10 Suppose three fair dice are roll... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose three fair dice are rolled. What is the probability at most one six appears?

Short Answer

Expert verified
The probability that at most one six appears when three fair dice are rolled is \(\frac{25}{27}\).

Step by step solution

01

Define events and complementary events

Let A be the event "at most one six appears." We are asked to find P(A). Let A' be the event "at least two sixes appear" (the complementary event). Then, we can express A as the complement of A': P(A) = 1 - P(A').
02

Calculate the probability of each outcome for a single dié

Since each die is fair and has six sides, the probability of rolling a six is 1/6 and the probability of not rolling a six is 5/6.
03

Compute the probability of A' using combinations

A' can occur when either two or three sixes appear. We can calculate the probability of these situations and sum them: a) Two sixes and one non-six: There are \(\binom{3}{2}\) = 3 ways to choose which two dice will have the sixes. The probability of this configuration is \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\). b) All three sixes: There is only one way that all three dice can have sixes, which has probability \(\left(\frac{1}{6}\right)^3\).
04

Compute P(A')

Now let's sum the probabilities from Step 3 to find P(A'): P(A') = 3 * \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\) + \(\left(\frac{1}{6}\right)^3\).
05

Compute P(A) using the complement

Now that we have P(A'), we can compute P(A) using the complement relationship: P(A) = 1 - P(A') = 1 - [3 * \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\) + \(\left(\frac{1}{6}\right)^3\)].
06

Evaluate and simplify

Evaluate and simplify the expression from Step 5: P(A) = 1 - [3 * \(\left(\frac{1}{36}\right)\left(\frac{5}{6}\right)\) + \(\frac{1}{216}\)] = 1 - [\(\frac{5}{72}\) + \(\frac{1}{216}\)]. First, convert the fractions to a common denominator (the least common multiple of 72 and 216 is 216): P(A) = 1 - [\(\frac{15}{216}\) + \(\frac{1}{216}\)]. Now, we can simplify the expression by combining the fractions: P(A) = 1 - \(\frac{16}{216}\). Simplify further by dividing by the greatest common divisor (8): P(A) = 1 - \(\frac{2}{27}\). Finally, convert the mixed number into an improper fraction: P(A) = \(\frac{27 - 2}{27}\) = \(\frac{25}{27}\).
07

State the final answer

So, the probability that at most one six appears when three fair dice are rolled is \(\frac{25}{27}\).

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!

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, each having a uniform distribution over \((0,1)\). Let \(M=\) maximum \(\left(X_{1}, X_{2}, \ldots, X_{n}\right)\). Show that the distribution function of \(M, F_{M}(\cdot)\), is given by $$ F_{M}(x)=x^{n}, \quad 0 \leq x \leq 1 $$ What is the probability density function of \(M ?\)

Suppose that an experiment can result in one of \(r\) possible outcomes, the ith outcome having probability \(p_{i}, i=1, \ldots, r, \sum_{i=1}^{r} p_{i}=1 .\) If \(n\) of these experiments are performed, and if the outcome of any one of the \(n\) does not affect the outcome of the other \(n-1\) experiments, then show that the probability that the first outcome appears \(x_{1}\) times, the second \(x_{2}\) times, and the \(r\) th \(x_{r}\) times is $$ \frac{n !}{x_{1} ! x_{2} ! \ldots x_{r} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{r}^{x_{r}} \quad \text { when } x_{1}+x_{2}+\cdots+x_{r}=n $$ This is known as the multinomial distribution.

Suppose a die is rolled twice. What are the possible values that the following random variables can take on? (a) The maximum value to appear in the two rolls. (b) The minimum value to appear in the two rolls. (c) The sum of the two rolls. (d) The value of the first roll minus the value of the second roll.

If \(X\) is a nonnegative random variable, and \(g\) is a differential function with \(g(0)=0\), then $$ E[g(X)]=\int_{0}^{\infty} P(X>t) g^{\prime}(t) d t $$ Prove the preceding when \(X\) is a continuous random variable.

If the coin is assumed fair, then, for \(n=2\), what are the probabilities associated with the values that \(X\) can take on?

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.