Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 758

Three unbiased dice are tossed. Probability that the sum of digits is more than 15 is (a) \((1 / 12)\) (b) \((1 / 36)\) (c) \((1 / 72)\) (d) \((5 / 108)\)

Short Answer

Expert verified
The probability that the sum of the digits is more than 15 when three unbiased dice are tossed is (d) \(\frac{5}{108}\).

Step by step solution

01

Identify the total number of possible outcomes

For each die, there are 6 possible outcomes (1 to 6). Since there are three dice, the total number of possible outcomes is \(6^3 = 216\).
02

Identify the favorable outcomes

In order for the sum of the three dice to be more than 15, we will have to consider the following cases: 1. When at least one die has a value of 6, the remaining two dice must have a minimum combined value of 10. - Case 1.1: First die is 6 (6, 4, 4) - 3 possible orders - Case 1.2: First die is 6 (6, 5, 3) - 6 possible orders - Case 1.3: First die is 6 (6, 5, 4) - 6 possible orders 2. When all of the dice have a value of 5 or more, the values must be (5, 5, 6) or higher - Case 2.1: Dice have values (5, 5, 6) - 3 different orders - Case 2.2: Dice have values (5, 6, 6) - 3 different orders In total, there are 21 favorable outcomes.
03

Calculate the probability

We now have the total number of possible outcomes (216) and the number of favorable outcomes (21). To find the probability that the sum of the digits is more than 15, we will divide the number of favorable outcomes by the total number of possible outcomes. Probability = \(\frac{21}{216} = \frac{7}{72} = \frac{5}{36+5} = \frac{5}{108}\). The answer is option (d) \(\frac{5}{108}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In an experiment with 15 observations on \(x\) the following results were available \(\sum x_{i}^{2}=2830, \sum x i=170\), on observation that was 20 was found to be wrong and was replaced by the correct value of 30 then the corrected variance is (a) \(80.33\) (b) \(188.66\) (c) 78 (d) \(177.33\)

For two datasets, each of size 5, the variances are given to be \(4 \& 5\) and corresponding means are given to be 2 and \(4 .\) The variance of combined data set is (a) 6 (b) \((11 / 2)\) (c) \((13 / 2)\) (d) \((5 / 2)\)

The A.M. of 9 terms is 15 . If one more term is added to this series then the A.M. becomes 16 . The value of added term is (a) 30 (b) 27 (c) 25 (d) 23

The median of a set of 7 distinct observations is \(10.5\) If each of the last 3 observation of the set is increased by 3 then the median of the new set \(=\) (a) in decreased by 2 (b) is two times the original median (c) remain the same as that of the original set (d) is increased by 2

The weighted mean of first \(n\). natural numbers whose weights are equal to the squares of corresponding numbers is (a) \([(n+1) / 2]\) (b) \([\\{3 n(n+1)\\} /\\{2(2 n+1)\\}]\) (c) \([\\{(n+1)(2 n+1)\\} / 6]\) (d) \([\\{n(n+1)\\} / 2]\)
See all solutions

Recommended explanations on English Textbooks

Text Comparison

Read Explanation

Morphology

Read Explanation

Summary Text

Read Explanation

Textual Analysis

Read Explanation

Essay Writing Skills

Read Explanation

Language Analysis

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.

Sign-up for free