/*! 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 30 In Exercises 25 - 30, find the p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 25 - 30, find the probability for the experiment of tossing a six-sided die twice. The sum is odd or prime

Short Answer

Expert verified
The probability that the sum is odd or prime when a six-sided dice is thrown twice is \(\frac{1}{2}\) or 0.5.

Step by step solution

01

All Possible Outcomes

Firstly, calculate the total possible outcomes when tossing a six-sided die twice. This is calculated as 6 (from the 1st die) multiplied by 6 (from the 2nd die), so there are \(6 \times 6 = 36\) total outcomes.
02

Outcomes that Sum to Odd Numbers

Secondly, calculate the outcomes that sum to odd numbers. An odd number can be obtained by adding an even and an odd number. When a die is thrown, it can show 3 odd (1,3,5) and 3 even numbers (2,4,6). Thus, the combinations that give odd sums when two dice are thrown is \(3 \times 3 = 9\) (odd from die one and even from die two) and \(3 \times 3 = 9\) (even from die one and odd from die two). So, there are a total of \(9+9 = 18\) outcomes with an odd sum.
03

Outcomes that Sum to Prime Numbers

Finally, calculate the outcomes that sum to prime numbers. The prime numbers less than 12 are 2, 3, 5, 7, 11. Carefully count the number of outcomes for each prime number. Sum is 2: (only possible with (1,1) so 1 outcome). Sum is 3: (1,2 and 2,1 or 2 outcomes). Sum is 5: (1,4; 2,3; 3,2; 4,1; 1,4; and 4,1 or 4 outcomes). Sum is 7: (1,6; 2,5; 3,4; 4,3; 5,2; 6,1-6 outcomes). Sum is 11: (5,6 and 6,5-2 outcomes). That gives a total of \(1 + 2 + 4 + 6 + 2 = 15\) outcomes that give a prime number sum.
04

Calculate the Probability

The final step is to calculate the probability. We already know that 18 outcomes constitute an odd sum, as calculated in step 2. We also know that among these 18 outcomes, 15 are also prime (as calculated in step 3). Therefore, all possible outcomes of interest are 18, as the group 'odd sum' encompasses all the outcomes of 'prime sum'. Now, to calculate the probability that the sum is odd or prime, we divide the number of outcomes of interest (18) by the total number of possible outcomes (36): \(\frac{18}{36}\).

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.

Prime Numbers
Prime numbers are fascinating numbers in mathematics that have specific properties. A prime number is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers. That means it has no divisors other than 1 and itself.

Some of the most well-known smaller prime numbers include:
  • 2
  • 3
  • 5
  • 7
  • 11
  • 13
These numbers appear frequently in probability theory because of their unique characteristics.

In the context of tossing two dice, finding sums that equate to prime numbers can be intriguing. Each combination of dice rolls that leads to sums such as 2, 3, 5, 7, and 11 requires us to consider the specific rolls needed. This makes prime numbers distinct and interesting when calculating probabilities.
Odd Numbers
Odd numbers have a key feature: they are numbers that cannot be divided evenly by 2. When dealing with dice, odd numbers often come into play. For a single die, the odd numbers are 1, 3, and 5.

When we toss two dice and look for sums that are odd, we rely on basic arithmetic. An odd sum occurs when one die shows an even number, and the other shows an odd number. This combination ensures the addition will result in an odd number.

So, for instance, the first die showing a 2 (even) and the second showing a 3 (odd), gives the odd sum of 5. Similarly:
  • 1 and 4 make 5.
  • 3 and 2 make 5.
The concept of odd numbers simplifies part of probability calculations, as nearly half of all number sums are odd when rolling two dice.
Probability Calculation
Probability calculation is the process of determining how likely an event is to occur. It's a fundamental part of probability theory and mathematics.

Common steps for probability calculation include:
  • Determine the total number of possible outcomes.
  • Identify the number of favorable outcomes.
  • Use the probability formula: \( \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Possible Outcomes}} \)
Let's apply this to our dice example. First, calculate the total possible outcomes—here, it's 36 when tossing two dice. Then determine how many of these outcomes result in an odd or prime sum, which we found to be 18.

Using the formula, the probability would thus be:
\[ \frac{18}{36} = \frac{1}{2} \]

This tells us there's a 50% chance the sum of the dice will be odd or prime, making probability calculation both simple and insightful in understanding randomness and chance.

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 Exercises 73 - 76, find the number of diagonals of the polygon. (A line segment connecting any two non adjacent vertices is called a diagonal of the polygon.) Decagon (\( 10 \) sides)

American roulette is a game in which a wheel turns on a spindle and is divided into \( 38 \) pockets.Thirty-six of the pockets are numbered \( 1-36 \), of which half are red and half are black. Two of the pockets are green and are numbered \( 0 \) and \( 00 \) (see figure). The dealer spins the wheel and a small ball in opposite directions. As the ball slows to a stop, it has an equal probability of landing in any of the numbered pockets. (a) Find the probability of landing in the number \( 00 \) pocket. (b) Find the probability of landing in a red pocket. (c) Find the probability of landing in a green pocket or a black pocket. (d) Find the probability of landing in the number \( 14 \) pocket on two consecutive spins. (e) Find the probability of landing in a red pocket on three consecutive spins.

In Exercises 39 - 42, you are given the probability that an event will not happen. Find the probability that the event will happen. \( P(E') = \dfrac{61}{100} \)

Write a short paragraph defining the following. (a) Sample space of an experiment (b) Event (c) The probability of an event \( E \) in a sample space \( S \) (d) The probability of the complement of \( E \)

In Exercises 53 - 60, the sample spaces are large and you should use the counting principles discussed in Section 9.6. One card is selected at random from an ordinary deck of \( 52 \) playing cards. Find the probabilities that (a) the card is an even-numbered card,(b) the card is a heart or a diamond, and (c) the card is a nine or a face card.

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.