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What is the probability that the sum of the numbers on two dice is even when they are rolled?

Short Answer

Expert verified
The probability that the sum is even is \(\frac{1}{2}\).

Step by step solution

01

Understand the Problem

To find the probability that the sum of the numbers on two dice is even, first recall that a sum is even if both numbers are even or both numbers are odd.
02

Determine Possible Outcomes

Each die has 6 faces, so there are a total of 6 × 6 = 36 possible outcomes when two dice are rolled.
03

Identify Favorable Outcomes

List the combinations where the sum is even (both even or both odd): (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) and (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5). Each of these pairs contributes to an even sum. There are 9 even-even combinations and 9 odd-odd combinations.
04

Count Favorable Outcomes

Since both even-even and odd-odd contribute to even sums: 9 (even-even) + 9 (odd-odd) = 18 favorable outcomes.
05

Calculate Probability

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. Therefore, the probability is \(\frac{18}{36} = \frac{1}{2}\).

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Key Concepts

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

even and odd sums

To understand the concept of even and odd sums, we need to first grasp what makes a number even or odd. An even number is divisible by 2 without any remainder (e.g., 2, 4, 6), while an odd number will have a remainder of 1 when divided by 2 (e.g., 1, 3, 5). When adding two numbers:

  • Two even numbers always result in an even sum. For example, 2 + 4 = 6.
  • Two odd numbers also result in an even sum. For example, 1 + 3 = 4.
  • An even and an odd number together result in an odd sum. For example, 2 + 3 = 5.

So, for the sum of numbers on two dice to be even, both rolled numbers must be either even or both must be odd.

dice probability

Probability is a measure of how likely an event is to occur. When rolling two dice, each die has 6 faces, resulting in a total of 6 × 6 = 36 possible outcomes. Each combination is equally likely. Knowing the total number of outcomes helps us calculate the likelihood of specific events, such as rolling certain sums.

When examining the probability of obtaining an even sum with two dice, we consider the individual probabilities of rolling each face. Each die can show a number between 1 and 6. Since these outcomes are independent of one another, we multiply their probabilities when looking at combined events.

For understanding how probability works in simpler terms, think of the steps as follows:

  • Count total possible outcomes (rolling two dice gives 36).
  • Identify and count favorable outcomes (18 favorable pairs where the sum is even).
  • Calculate the probability by dividing favorable outcomes by total outcomes: \(\frac{18}{36} = \frac{1}{2} \).
favorable outcomes

In probability, favorable outcomes are those that satisfy the condition of the event we are interested in. For the problem at hand, the event is getting an even sum from rolling two dice. Favorable outcomes include:

  • Pairs where both numbers are even: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), and (6,6) - 9 combinations.
  • Pairs where both numbers are odd: (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), and (5,5) - 9 combinations.

Combining these, we find 18 favorable outcomes. Since these are the combinations leading to an even sum, the probability of rolling an even sum with two dice is determined by dividing these favorable outcomes by the total number of possible outcomes, which is 36:

This results in a probability of \(\frac{18}{36} = \frac{1}{2} \), meaning there is a 50% chance of rolling an even sum with two dice.

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Most popular questions from this chapter

Prove Theorem \(2,\) the extended form of Bayes' theorem. That is, suppose that \(E\) is an event from a sample space \(S\) and that \(F_{1}, F_{2}, \ldots, F_{n}\) are mutually exclusive events such that \(\bigcup_{i=1}^{n} F_{i}=S .\) Assume that \(p(E) \neq 0\) and \(p\left(F_{i}\right) \neq 0\) for \(i=1,2, \ldots, n .\) Show that $$ p\left(F_{j} | E\right)=\frac{p\left(E | F_{j}\right) p\left(F_{j}\right)}{\sum_{i=1}^{n} p\left(E | F_{i}\right) p\left(F_{i}\right)} $$ \(\left[\text {Hint} : \text { Use the fact that } E=\bigcup_{i=1}^{n}\left(E \cap F_{i}\right) .\right]\)

A pair of dice is loaded. The probability that a 4 appears on the first die is \(2 / 7,\) and the probability that a 3 appears on the second die is \(2 / 7 .\) Other outcomes for each die appear with probability \(1 / 7 .\) What is the probability of 7 appearing as the sum of the numbers when the two dice are rolled?

In this exercise we will use Bayes’ theorem to solve the Monty Hall puzzle (Example 10 in Section 7.1). Recall that in this puzzle you are asked to select one of three doors to open. There is a large prize behind one of the three doors and the other two doors are losers. After you select a door, Monty Hall opens one of the two doors you did not select that he knows is a losing door, selecting at random if both are losing doors. Monty asks you whether you would like to switch doors. Suppose that the three doors in the puzzle are labeled 1, 2, and 3. Let W be the random variable whose value is the number of the winning door; assume that p(W = k) = 1?3 for k = 1, 2, 3. Let M denote the random variable whose value is the number of the door that Monty opens. Suppose you choose door i. a) What is the probability that you will win the prize if the game ends without Monty asking you whether you want to change doors? b) Find p(M = j ? W = k) for j = 1, 2, 3 and k = 1, 2, 3. c) Use Bayes’ theorem to find p(W = j ? M = k) where i and j and k are distinct values. d) Explain why the answer to part (c) tells you whether you should change doors when Monty gives you the chance to do so.

In roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and \(00 .\) The probability that when the wheel is spun it lands on any particular number is 1\(/ 38\) . a) What is the probability that the wheel lands on a red number? b) What is the probability that the wheel lands on a black number twice in a row? c) What is the probability that the wheel lands on 0 or 00\(?\) do? d) What is the probability that in five spins the wheel never lands on either 0 or 00\(?\) e) What is the probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin?

Two events \(E_{1}\) and \(E_{2}\) are called independent if \(p\left(E_{1} \cap E_{2}\right)=p\left(E_{1}\right) p\left(E_{2}\right) .\) For each of the following pairs of events, which are subsets of the set of all possible outcomes when a coin is tossed three times, determine whether or not they are independent. a) \(E_{1} :\) tails comes up with the coin is tossed the first time; \(E_{2} :\) heads comes up when the coin is tossed the second time. b) \(E_{1} :\) the first coin comes up tails; \(E_{2} :\) two, and not three, heads come up in a row. c) \(E_{1} :\) the second coin comes up tails; \(E_{2} :\) two, and not three, heads come up in a row. (We will study independence of events in more depth in Section \(7.2 . )\)

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