/*! 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 3 Two dice are thrown. Let \(E\) b... [FREE SOLUTION] | 91Ó°ÊÓ

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Two dice are thrown. Let \(E\) be the event that the sum of the dice is odd; let \(F\) be the event that at least one of the dice lands on 1 ; and let \(G\) be the event that the sum is 5 . Describe the events \(E F, E \cup F, F G, E F^{c}\), and \(E F G\).

Short Answer

Expert verified
The events \(E \cap F, E \cup F, F \cap G, E \cap F^{c},\) and \(E \cap F \cap G\) contain the following outcomes: \(E \cap F = \{(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)\}\) \(E \cup F = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5) \}\) \(F \cap G = \{(4, 1)\}\) \(E \cap F^{c} = \{(2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 3), (6, 5)\}\) \(E \cap F \cap G = \emptyset\)

Step by step solution

01

1. Event E (sum of the dice is odd)

Event E consists of all the outcomes in which the sum of the numbers on the two dice rolled is odd. To list the outcomes in E, we will consider every combination in the sample space and select those with an odd sum: \(E = \{(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)\}\)
02

2. Event F (at least one die lands on 1)

Event F consists of all the outcomes in which at least one of the dice rolled shows a 1: \(F = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)\}\)
03

3. Event G (sum of the dice is 5)

Event G consists of all the outcomes in which the sum of the numbers on the two dice rolled is 5: \(G = \{(1, 4), (2, 3), (3, 2), (4, 1)\}\) Now, we will describe the requested intersections and unions:
04

4. Event \(E \cap F\) (sum is odd and at least one die lands on 1)

This event includes the outcomes that are both in events E and F: \(E \cap F = \{(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)\}\)
05

5. Event \(E \cup F\) (sum is odd or at least one die lands on 1, or both)

This event includes all the outcomes that are in either E or F, or both: \(E \cup F = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)\}\)
06

6. Event \(F \cap G\) (at least one die lands on 1 and sum of the dice is 5)

This event includes the outcomes that are both in events F and G: \(F \cap G = \{(4, 1)\}\)
07

7. Event \(E \cap F^{c}\) (sum is odd and neither die lands on 1)

First, let's describe the complement of F, or the outcomes in which neither die lands on 1: \(F^{c} = \{(2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}\) Now, to find \(E \cap F^{c}\), we will list the outcomes that are both in events E and \(F^{c}\): \(E \cap F^{c} = \{(2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 3), (6, 5)\}\)
08

8. Event \(E \cap F \cap G\) (sum is odd, at least one die lands on 1, and sum is 5)

This event includes the outcomes that are in events E, F, and G at the same time. Since G has only outcomes with an even sum, and E has only outcomes with an odd sum, these events are mutually exclusive, which means there can be no common outcomes: \(E \cap F \cap G = \emptyset\)

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Key Concepts

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

Events and outcomes
In probability theory, an event is a set of outcomes from a specific experiment. Think of an event as a scenario that we are interested in. For example, when rolling two dice, we might be interested in the event where the sum of the numbers is odd. Events can be simple, containing just one outcome, or compound, containing multiple outcomes.
When you roll two dice:
  • A simple outcome could be rolling a (1, 2).
  • The event of having an odd sum (like event E in our exercise) includes several outcomes, such as (1, 2), (2, 3), or (5, 4), where each pair consists of a roll on each die.
Various outcomes combine to form events, shaping the basis of probability calculations.
Intersection and union of events
The intersection and union of events help to explore combined scenarios. The intersection, denoted as \(E \cap F\), refers to the outcomes that occur in both events simultaneously. For instance, if event E is having an odd sum and event F is at least one die landing on 1, then \(E \cap F\) includes outcomes like (1, 2), where both conditions are satisfied.
The union, represented by \(E \cup F\), includes outcomes that occur in either event or in both. So for events E and F, \(E \cup F\) consists of any outcome where the sum is odd or at least one die shows 1, like (1, 4) or (2, 1). These concepts help us to understand possibilities when we look at overlapping and inclusive combinations of events.
Complementary events
Complementary events include all possible outcomes that are not in the given event. For every event A, its complement \(A^c\) consists of all outcomes that do not belong to A. This plays a crucial role when considering what something is not in probability.
For instance, if event F is that at least one die lands on 1, then \(F^c\)—the complement—includes all outcomes where neither die shows a 1, such as (2, 2), (3, 4), and so on.
This idea is especially useful to find probabilities indirectly, as often, it can be easier to calculate the probability of the complement and subtract it from 1, giving us the probability of the original event.
Sample space and sample points
A sample space is the set of all possible outcomes of an experiment. Each possible outcome in the sample space is called a sample point. When throwing two dice, the sample space includes 36 sample points, considering all combinations from (1, 1) to (6, 6).
  • The sample space helps to ensure that we're considering every potential scenario.
  • It provides a comprehensive view from which events can be formed.
Understanding the full sample space and its sample points is essential as it lays the groundwork for calculating probabilities. It allows us to visualize and correctly enumerate the condition-specific outcomes that define each event.

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Most popular questions from this chapter

The following data were given in a study of a group of 1000 subscribers to a certain magazine: In reference to job, marital status, and education, there were 312 professionals, 470 married persons, 525 college graduates, 42 professional college graduates, 147 married college graduates, 86 married professionals, and 25 married professional college graduates. Show that the numbers reported in the study must be incorrect. HINT: Let \(M, W\), and \(G\) denote, respectively, the set of professionals, married persons, and college graduates. Assume that one of the 1000 persons is chosen at random and use Proposition \(4.4\) to show that if the numbers above are correct, then \(P(M \cup W \cup G)>1\).

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The game of craps is played as follows: A player rolls two dice. If the sum of the dice is either a 2,3 , or 12 , the player loses; if the sum is either a 7 or an 11 , he or she wins. If the outcome is anything else, the player continues to roll the dice until he or she rolls either the initial outcome or a 7. If the 7 comes first, the player loses; whereas if the initial outcome reoccurs before the 7, the player wins. Compute the probability of a player winning at craps. HinT: Let \(E_{i}\) denote the event that the initial outcome is \(i\) and the player wins. The desired probability is \(\sum_{i=2}^{12} P\left(E_{i}\right) .\) To compute \(P\left(E_{i}\right)\), define the events \(E_{L, n}\) to be the event that the initial sum is \(i\) and the player wins on the \(n\)th roll. Argue that \(P\left(E_{i}\right)=\sum_{n=1}^{\infty} P\left(E_{i, n}\right)\).

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