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Chapter 7: Problem 38

A pair of dice is rolled in a remote location and when you ask an honest observer whether at least one die came up six, this honest observer answers in the affirmative. a) What is the probability that the sum of the numbers that came up on the two dice is seven, given the information provided by the honest observer? b) Suppose that the honest observer tells us that at least one die came up five. What is the probability the sum of the numbers that came up on the dice is seven, given this information?

Short Answer

Expert verified
a) \( \frac{2}{11} \); b) \( \frac{2}{11} \).

Step by step solution

01

Identify the Number of Possible Outcomes

When rolling two dice, there are a total of 36 possible outcomes (6 outcomes per die).
02

Count Favorable Outcomes for At Least One Die Being Six

For at least one die to show a six, there are 11 possible outcomes: (1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2), (6,3), (6,4), (6,5).
03

Count Favorable Outcomes Where Sum is Seven with One Die Showing Six

The only outcome in which at least one die is six and the sum is seven is (1,6) and (6,1). There are 2 favorable outcomes.
04

Calculate Conditional Probability for Part a

The probability that the sum is seven given that at least one die shows six is: \[ \text{P(sum is 7 | at least one 6)} = \frac{\text{Number of favorable outcomes}}{\text{Number of outcomes with at least one 6}} = \frac{2}{11} \]
05

Count Outcomes for At Least One Die Being Five

For at least one die to show a five, there are also 11 possible outcomes: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6).
06

Count Favorable Outcomes Where Sum is Seven with One Die Showing Five

The only outcomes in which at least one die is five and the sum is seven is (2,5) and (5,2). There are 2 favorable outcomes.
07

Calculate Conditional Probability for Part b

The probability that the sum is seven given that at least one die shows five is: \[ \text{P(sum is 7 | at least one 5)} = \frac{\text{Number of favorable outcomes}}{\text{Number of outcomes with at least one 5}} = \frac{2}{11} \]

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random phenomena. At its core, it deals with the likelihood of different outcomes. In probability theory:
  • An **experiment** is a process or action which results in one or more outcomes. In our exercise, rolling two dice is an experiment.
  • **Outcomes** are possible results of an experiment. For two dice, there are 36 unique outcomes (6 per die, resulting in 6 x 6 = 36).
  • A **sample space** is the set of all possible outcomes. For two dice, the sample space includes all pairs of rolls (1,1), (1,2), ..., (6,6).
  • An **event** is a subset of the sample space. For example, getting a sum of 7 when rolling two dice is an event.
To find the probability of an event, we calculate:
\( \text{P(Event)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \)
Outcomes
In any probability problem, identifying the number of possible outcomes is crucial. This helps us understand the total scenarios we're dealing with.
  • For two dice, there are 36 outcomes because each die has 6 faces, resulting in \( 6 \times 6 = 36 \) combinations.
  • Each outcome for the dice can be written as an ordered pair such as (1,1), (1,2), ..., (6,6).
  • Once we know there are 36 outcomes, we can count which of these are favorable for our events. In our exercise, these events are: 'at least one die shows six' and 'sum is seven'.
So, when counting the outcomes:
  • For 'at least one die shows six', our favorable outcomes include all pairs where at least one of the dice shows a 6: (1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2), (6,3), (6,4), and (6,5). These are 11 possibilities.
  • For 'sum is seven' with 'at least one die shows six', the only outcomes are (1,6) and (6,1), as these pairs add up to 7. These are 2 possibilities.
Counting these helps us to set up for calculating conditional probability.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. To express conditional probability, we use the notation \( P(A|B) \), which reads 'the probability of A given B'.
  • In our exercise, we need to find the probability of getting a sum of 7, given that at least one die shows 6.
  • This is written as \( P(\text{Sum is 7} | \text{At least one 6}) \).
  • The formula to calculate this is: \( P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \).
Let's apply it to the exercise:
  • From our previous steps, we know there are 11 possible outcomes where at least one die shows six. So, \( P(\text{At least one 6}) = \frac{11}{36} \).
  • The outcomes where the sum is 7 and one die is 6 are (1,6) and (6,1). So, \( P(\text{Sum is 7 and one 6}) = \frac{2}{36} \).
  • Thus, using the formula: \( P(\text{Sum is 7} | \text{At least one 6}) = \frac{2/36}{11/36} = \frac{2}{11} \).
Therefore, the probability of getting a sum of 7, given that one die is a 6, is \( \frac{2}{11} \).

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Most popular questions from this chapter

A player in the Powerball lottery picks five different integers between 1 and 69 , inclusive, and a sixth integer between 1 and \(26,\) which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn. a) What is the probability that a player wins the jackpot? b) What is the probability that a player wins \(\$ 1,000,000\) , which is the prize for matching the first five numbers, but not the sixth number, drawn? c) What is the probability that a player wins \(\$ 100\) by matching exactly three of the first five and the sixth numbers drawn, or four of the first five numbers, but not the sixth number, drawn? d) What is the probability that a player wins a prize of \(\$ 4,\) which is the prize when the player matches the sixth number, and either one or none of the first five numbers drawn?

What is the probability that when a coin is flipped six times in a row, it lands heads up every time?

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?

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If \(E\) and \(F\) are independent events, prove or disprove that \(\overline{E}\) and \(F\) are necessarily independent events.
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