/*! 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 9 Two fair dice are rolled. Show t... [FREE SOLUTION] | 91Ó°ÊÓ

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Two fair dice are rolled. Show that the event that their sum is 7 is independent of the score shown. by the first die.

Short Answer

Expert verified
The event of the sum being 7 is independent of the first die.

Step by step solution

01

Identify Possible Outcomes

When two dice are rolled, each die has 6 possible outcomes. Thus, there are a total of \(6 \times 6 = 36\) possible pairs of outcomes \((i, j)\) where \(i\) and \(j\) are the results of the first and second die, respectively.
02

Determine the Event Sum is 7

Identify the pairs \((i, j)\) where the sum is 7: \((1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\). There are 6 pairs.
03

Determine the Probability of Sum is 7

The probability of the sum being 7 is given by the formula \(P(S = 7) = \frac{\text{number of favorable outcomes}}{\text{total outcomes}} = \frac{6}{36} = \frac{1}{6}\).
04

Determine Probability Based on the First Die

Calculate the probability of the event that the first die shows a specific number. Since all numbers are equally likely, \(P(A_i) = \frac{1}{6}\) for any number \(i\) from 1 to 6.
05

Determine Conditional Probability

Calculate the probability of the sum being 7 given the first die shows a specific number \(i\). For each possible \(i\), there is exactly one outcome \(j\) such that \(i + j = 7\). Thus, \(P(S = 7 \mid A_i) = \frac{1}{6}\).
06

Establish Independence

An event is independent if \(P(B \mid A) = P(B)\). Here, \(P(S = 7 \mid A_i) = P(S = 7) = \frac{1}{6}\) for any result \(i\). Hence, the event of the sum being 7 is independent of the number shown on the first die.

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Key Concepts

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

Independence of Events
In probability theory, the independence of events is a core concept that helps us understand how events interact with each other. Two events are considered independent if the occurrence of one does not affect the probability of the other happening.
For instance, when rolling two dice, each roll operates independently of the other. This means that the outcome of the first die does not influence the outcome of the second.
Additionally, independence can be mathematically tested by checking if the probability of one event given the other is the same as the probability of the event alone. For example, if we find that the probability of getting a sum of 7 when the first die shows a specific number is the same as the probability of rolling a 7 regardless of the first die, then these two events are independent. In the exercise, this is confirmed by showing that for every possible roll on the first die, the probability of a total sum of 7 remains constant.
Conditional Probability
Conditional probability focuses on the likelihood of an event occurring given that another event has already occurred.
Mathematically, it's represented as \(P(A \mid B)\), which means the probability of event \(A\) happening given that \(B\) has already happened. Using the exercise as an example, we examine the probability of the sum of the dice being 7, assuming we know the result of the first die.
  • Calculate the probability of a specific outcome of interest (sum equals 7) relying on known information (first die roll).
  • Revenue is generated by filtering the relevant possibilities based on given constraints, such as the first die's outcome.
Understanding conditional probability allows us to see how certain information can change the calculation of outcomes. However, in the solution for our specific problem, we found the conditional probability was the same as the overall probability of rolling a sum of 7, helping us identify independence.
Dice Probability
Dice probability forms the basis for understanding simple probabilistic events in games of chance. When a fair six-sided die is rolled, each face has an equal probability of landing face up, which is \(\frac{1}{6}\).
When two dice are rolled, the possible outcomes include every combination of numbers from both dice, creating a complete set of 36 possible outcomes. Analyzing these outcomes allows us to calculate the probabilities of various sums, like rolling a total of 7. This calculation includes listing all outcome pairs that achieve the sum.
In our exercise, the probability of rolling a sum of 7 was computed by identifying all the pairs of numbers that add up to 7, with each pair representing an equally likely dice outcome. Working through dice probability problems helps students apply fundamental principles of probability to practical, interesting scenarios.
Event Outcomes
Event outcomes represent the potential results of a probabilistic event, such as rolling dice.
For two dice, each roll results in an outcome pair \((i, j)\), where \(i\) is the result on the first die and \(j\) on the second. The number of potential outcomes for two dice is 36, calculated as 6 outcomes from the first die multiplied by 6 from the second.
Identifying possible outcomes is the first step in evaluating probabilities. By understanding all potential event outcomes, we can determine the likelihood of specific events, such as rolling a sum of 7 in our exercise. The ability to list and analyze these outcomes enhances our understanding of larger probability questions and supports the calculation of probabilities related to other events. Through practice, calculating event outcomes sharpens problem-solving and critical-thinking skills, essential for mastering probability theory.

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Most popular questions from this chapter

A fair coin is tossed repeatedly. Show that the following two statements are equivalent: (a) the outcomes of different tosses are independent. (b) for any given finite sequence of heads and tails, the chance of this sequence occurring in the first \(m\) tosses is \(2^{-m}\), where \(m\) is the length of the sequence.

Let \(A_{r}, r \geq 1\), be events such that \(P\left(A_{r}\right)=1\) for all \(r\). Show that \(\mathbb{P}\left(\bigcap_{r=1}^{\infty} A_{r}\right)=1\).

For events \(A_{1}, A_{2} \ldots . A_{n}\) satisfying \(\mathrm{P}\left(A_{1} \cap A_{2} \cap \cdots \cap A_{n-1}\right)>0\), prove that $$ \mathrm{P}\left(A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right)=\mathbb{P}\left(A_{1}\right) \mathbb{P}\left(A_{2} \mid A_{1}\right) P\left(A_{3} \mid A_{1} \cap A_{2}\right) \cdots P\left(A_{n} \mid A_{1} \cap A_{2} \cap \cdots \cap A_{n-1}\right) $$

A symmetric random walk takes place on the integers \(0,1,2, \ldots . N\) with absorbing barriers at 0 and \(N\), starting at \(k\). Show that the probability that the walk is never absorbed is zero.

There are two roads from \(\mathrm{A}\) to \(\mathrm{B}\) and two roads from \(\mathrm{B}\) to \(\mathrm{C}\). Each of the four roads is blocked by snow with probability \(p\), independently of the others. Find the probability that there is an open road from \(\mathrm{A}\) to \(\mathrm{B}\) given that there is no open route from \(\mathrm{A}\) to \(\mathrm{C}\). If, in addition, there is a direct road from \(\mathrm{A}\) to \(\mathrm{C}\), this road being blocked with probability \(p\). independently of the others, find the required conditional probability.

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