/*! 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 90 A fair die is rolled three times... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A fair die is rolled three times. Let \(X\) denote the number of different faces showing, \(X=1,2,3\). Find \(E(X)\).

Short Answer

Expert verified
The expected value is the sum of the products of each outcome (number of different faces) and its probability.

Step by step solution

01

Identify Possible Outcomes For X

First, determine the number of ways each number of different faces can occur. That is, find the number of ways one can get 1, 2, or 3 different faces when rolling a die three times.
02

Calculate The Probability For Each Outcome

Next, calculate the probability of each of these options happening. The total number of outcomes when a die is rolled three times is \(6^3\). Use this to calculate the probabilities by dividing the number of ways each number of different faces can occur by the total number of outcomes.
03

Calculate Expected Value

The last step is to calculate the expected value. The expected value \(E(X)\) for a discrete random variable is found by summing the product of each outcome and its likelihood \(E(X) = Σ [xi * P(xi)]\).

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Key Concepts

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

Probability Theory
Probability theory helps us understand how likely events are to happen. When rolling a fair six-sided die three times, we're interested in how many different faces show up. Each roll has 6 possible outcomes, so rolling three times gives us a total of \(6^3 = 216\) possible outcomes.
For any event, probability is determined by dividing the number of ways an event can happen by the number of all possible outcomes. For example, the probability of rolling a 1 on a die is \(\frac{1}{6}\) because there is only one way to roll a 1 and six possible outcomes. When you know probabilities, you can start to predict outcomes more reliably, like determining the expected number of different faces in our die-rolling exercise.
In our case, we calculate probabilities for having 1, 2, or 3 different face outcomes and use it for further analysis, like calculating the expected value.
Discrete Random Variables
A discrete random variable is a type of variable that can take on only a specific set of values. In our die-rolling example, the variable \(X\) represents the number of different faces showing, and because a die has six faces, \(X\) can only be 1, 2, or 3.
This restriction is what makes \(X\) a discrete random variable. You can't have 1.5 different faces showing; the outcome must be an integer. Discrete random variables can take on values in a list, not a range. Such variables are essential in calculating probabilities because they simplify situations involving counting and distinct outcomes.
Understanding discrete random variables allows us to solve for things like expected value using a clear, systematic approach, making it simpler to forecast aspects of chance-driven systems.
Combinatorial Analysis
Combinatorial analysis is a method used to count how many different ways certain things can happen, which is crucial in calculating probabilities. In the dice exercise, we used combinatorial analysis to determine the number of ways to get 1, 2, or 3 distinct faces from three dice rolls.
To find these numbers, think about your restrictions:
  • Getting 1 face means all dice show the same number, which can happen in 6 ways (one for each possible number).
  • Getting 2 different faces involves picking 2 distinct numbers and arranging three dice to show these numbers, like two of one number and one of another.
  • Getting 3 different faces requires all dice to show different numbers, and selecting these numbers and arranging them in 3 positions is even more complex.
Combinatorial analysis enables this sort of counting, which then allows us to find the probability and ultimately the expected value. It breaks down complex scenarios into manageable parts, making it simpler to understand and solve probability exercises.

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Most popular questions from this chapter

A manufacturer has one hundred memory chips in stock, \(4 \%\) of which are likely to be defective (based on past experience). A random sample of twenty chips is selected and shipped to a factory that assembles laptops. Let \(X\) denote the number of computers that receive faulty memory chips. Find \(E(X)\).

. Suppose that \(X\) and \(Y\) are discrete random variables with $$ \begin{aligned} p_{X, Y}(x, y)=& \frac{4 !}{x ! y !(4-x-y) !}\left(\frac{1}{2}\right)^{x}\left(\frac{1}{3}\right)^{y}\left(\frac{1}{6}\right)^{4-x-y}, \\\ & 0 \leq x+y \leq 4 \end{aligned} $$ Find \(p_{X}(x)\) and \(p_{Y}(x)\).

A point is chosen at random from the interior of a circle whose equation is \(x^{2}+y^{2} \leq 4\). Let the random variables \(X\) and \(Y\) denote the \(x\) - and \(y\)-coordinates of the sampled point. Find \(f_{X, Y}(x, y)\).

. Suppose \(U\) is a uniform random variable over \([0,1]\). (a) Show that \(Y=(b-a) U+a\) is uniform over \([a, b]\). (b) Use part (a) and Theorem \(3.6 .2\) to find the variance of \(Y\).

Suppose a life insurance company sells a \(\$ 50,000\), five-year term policy to a twenty-five-year-old woman. At the beginning of each year the woman is alive, the company collects a premium of \(\$ P\). The probability that the woman dies and the company pays the \(\$ 50,000\) is given in the table below. So, for example, in Year 3 , the company loses \(\$ 50,000-\$ P\) with probability \(0.00054\) and gains \(\$ P\) with probability \(1-0.00054=0.99946\). If the company expects to make \(\$ 1000\) on this policy, what should \(P\) be? \begin{tabular}{cc} \hline Year & Probability of Payoff \\ \hline 1 & \(0.00051\) \\ 2 & \(0.00052\) \\ 3 & \(0.00054\) \\ 4 & \(0.00056\) \\ 5 & \(0.00059\) \\ \hline \end{tabular}

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