/*! 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 426 Find the probability of throwing... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 426

Find the probability of throwing two sixes in one toss of a pair of dice.

Short Answer

Expert verified
The probability of throwing two sixes in one toss of a pair of dice is \(\frac{1}{36}\) or approximately 2.78%.

Step by step solution

01

Determine the sample space (possible outcomes)

When tossing a pair of dice, there are 6 possible outcomes for the first die and 6 possible outcomes for the second die. So, there are 6 脳 6 = 36 possible outcomes in total.
02

Identify the event of interest (two sixes)

The event we are interested in, i.e. throwing two sixes, occurs only once when both dice show six. Therefore, there is 1 favorable outcome for our event of interest.
03

Calculate the probability

The probability of an event happening is the ratio of the number of favorable outcomes for the event to the total number of possible outcomes. In this case, the probability of throwing two sixes in one toss of a pair of dice can be calculated as: \[ P(\text{two sixes}) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}} = \frac{1}{36} \]
04

Express the probability as a fraction or a percentage

The probability of throwing two sixes in one toss of a pair of dice is \(\frac{1}{36}\). To convert this fraction to a percentage, simply multiply the fraction by 100: \[ \frac{1}{36} \times 100 \% \approx 2.78 \% \] So, the probability of throwing two sixes in one toss of a pair of dice is approximately 2.78%.

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 91影视!

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

A box contains 4 black marbles, 3 red marbles, and 2 white marbles. What is the probability that a black marble, then a red marble, then a white marble is drawn without replacement?

Four cards are to be dealt successively, at random and without replacement, from an ordinary deck of playing cards. Find the probability of receiving a spade, a heart, a diamond, and a club, in that order.

In the St. Petersburg Community College, \(30 \%\) of the men and \(20 \%\) of the women are studying mathematics. Further, \(45 \%\) of the students are women. If a student selected at random is studying mathematics, what is the probability that the student is a woman?

A box contains 4 black marbles, 3 red marbles, and 2 white marbles. What is the probability that a black marble, then a red marble, then a white marble is drawn without replacement?

An electronic device contains two easily removed subassemblies, \(\mathrm{A}\) and \(\mathrm{B}\). If the device fails, the probability that it will be necessary to replace A is \(0.50\). Some failures of A will damage \(\mathrm{B}\). If A must be replaced, the probability that \(\mathrm{B}\) will also have to be replaced is \(0.70\). If it is not necessary to replace A, the probability that \(\mathrm{B}\) will have to be replaced is only \(0.10\). What percentage of all failures will you require to replace both \(\mathrm{A}\) and \(\mathrm{B}\) ?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
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.