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Chapter 2: Problem 71

Two fair dice are rolled. What is the probability that the number on the first die was at least as large as 4 given that the sum of the two dice was \(8 ?\)

Short Answer

Expert verified
The required probability is \( \frac{1}{12} \).

Step by step solution

01

Determining the Sample Space

First identify the total number of outcomes when two dice are rolled, which is \(6 \times 6 = 36\) total outcomes.
02

Identify the Favourable Outcomes

Identify the pairs that sum to 8, bearing in mind the number on the first die is at least 4. These are (4, 4), (5, 3), and (6, 2). Note that (3, 5) and (2, 6) are not counted because the first die can't be 3 or 2. So, we have 3 such cases.
03

Calculating the Probability

The probability is then calculated by dividing the number of favourable outcomes by the total number of outcomes. So the required probability is \( \frac{3}{36} = \frac{1}{12} \).
04

Simplifying the Result

The result \( \frac{1}{12} \) cannot be simplified any further, hence it is the answer.

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Key Concepts

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

Fair Dice
A fair die is a common six-sided cube with numbers from 1 to 6, each having an equal probability of appearing when you roll the die.
In the context of probability, a "fair" die ensures each side or face of the die has the same chance of landing face up.
This means each number has a probability of \(\frac{1}{6}\) of being rolled.
When two fair dice are used, like in our exercise, the total outcomes possible are the product of the outcomes for each die.
  • Since each die has 6 sides, the total number of outcomes when rolling two dice is \(6 \times 6 = 36\).
  • This forms the basis for our sample space when calculating probabilities with dice.
A fair dice setup lets us accurately explore probability questions because results are not biased by a weighted or unfair die.
Sample Space
The sample space in probability refers to the set of all possible outcomes of a random experiment.
For the roll of two fair dice, the sample space includes all combinations of the numbers 1 through 6 that can appear on the two dice.
Consider each die rolls a face among these numbers.
  • Theoretically, this generates a total of \(6 \times 6 = 36\) different pairs or outcomes.
  • For example, \((1,1), (1,2), (1,3)\ldots, (6,6)\) are some of the paired possibilities.
When solving probability problems, it's crucial to define the sample space correctly; it frames how probabilities are calculated.
In our exercise, the sample space is used to identify relevant desired outcomes such as pairs summing to a specific number.
Favourable Outcomes
In probability, favourable outcomes are specific outcomes within the sample space that satisfy the condition of the problem.
For example, if we're interested in the probability of rolling a sum of 8 with two dice, we focus on only those pairs of numbers.
  • For outcomes where the first die has a number at least 4, the pairs are \( (4,4), (5,3), \ and \ (6,2)\).
  • These are the desired outcomes under the given condition, excluding pairs like \( (3,5) \) where the first die is not 4 or greater.
To calculate the probability, divide the number of favourable outcomes by the total possible outcomes.
For this exercise, this fraction is \(\frac{3}{36} = \frac{1}{12}\), providing the likelihood of meeting both conditions: first die at least 4 and total of 8.

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Most popular questions from this chapter

Your favorite college football team has had a good season so far but they need to win at least two of their last four games to qualify for a New Year's Day bowl bid. Oddsmakers estimate the team's probabilities of winning each of their last four games to be \(0.60,0.50,0.40\), and \(0.70\), respectively. (a) What are the chances that you will get to watch your team play on Jan. 1 ? (b) Is the probability that your team wins all four games given that they have won at least three games equal to the probability that they win the fourth game? Explain. (c) Is the probability that your team wins all four games given that they won the first three games equal to the probability that they win the fourth game?

A study has shown that seven out of ten people will say "heads" if asked to call a coin toss. Given that the coin is fair, though, a head occurs, on the average, only five times out of ten. Does it follow that you have the advantage if you let the other person call the toss? Explain.

Suppose each of ten sticks is broken into a long part and a short part. The twenty parts are arranged into ten pairs and glued back together so that again there are ten sticks. What is the probability that each long part will be paired with a short part? (Note: This problem is a model for the effects of radiation on a living cell. Each chromosome, as a result of being struck by ionizing radiation, breaks into two parts, one part containing the centromere. The cell will die unless the fragment containing the centromere recombines with a fragment not containing a centromere.)

A coke hand in bridge is one where none of the thirteen cards is an ace or is higher than a 9 . What is the probability of being dealt such a hand?

Use Venn diagrams to suggest an equivalent way of representing the following events: (a) \(\left(A \cap B^{C}\right)^{C}\) (b) \(B \cup(A \cup B)^{C}\) (c) \(A \cap(A \cap B)^{C}\)
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