Math Problem Statement

An number cube (a far die) is rolled 3 times. For each roll, we are interested in whether the roll comes up even or odd. An outcome is represented by a string of the sort bee (meaning an odd number on the first roll, an even number on the second roll, and an even number on the third roll).

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For each outcome, let N be the random vacable counting the number of even rolls in each outcome. For example, if the outcome is 000, then N (000) = 0. Suppose that the random variatile X is defined in terms of V as follows: X-6N-2N-1. The values of X are given in the table below.

Outcome

CEE

000

000

000

200

000

oee

ooe

Value

of X13-133

33

3

Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row

Value x of X

PX-x)

Solution

To solve this problem, we need to calculate the values of the random variable XX based on the given outcomes and then determine the probability distribution of XX. Let's break it down step by step.

Step 1: Understanding the Problem

  • We are rolling a fair die (which has numbers from 1 to 6) three times.
  • For each roll, the outcome can be either even (E) or odd (O).
  • Each outcome can be represented as a string like "EEE" or "OEO", where E stands for an even result and O stands for an odd result.
  • We are interested in the random variable NN, which counts the number of even results in each outcome.

Given that the random variable XX is defined as: X=6N2N1X = 6N - 2N - 1 we first need to calculate the value of XX for each possible outcome.

Step 2: Calculate XX for All Possible Outcomes

  • There are 23=82^3 = 8 possible outcomes for three rolls since each roll can result in either an even or an odd number.

Let's list the possible outcomes and calculate NN and XX for each:

OutcomeNN (Number of Evens)X=6N2N1X = 6N - 2N - 1
OOO06(0)2(0)1=16(0) - 2(0) - 1 = -1
OOE16(1)2(1)1=36(1) - 2(1) - 1 = 3
OEO16(1)2(1)1=36(1) - 2(1) - 1 = 3
OEE26(2)2(2)1=76(2) - 2(2) - 1 = 7
EOO16(1)2(1)1=36(1) - 2(1) - 1 = 3
EOE26(2)2(2)1=76(2) - 2(2) - 1 = 7
EEO26(2)2(2)1=76(2) - 2(2) - 1 = 7
EEE36(3)2(3)1=116(3) - 2(3) - 1 = 11

Step 3: Determine the Probability Distribution of XX

Now we need to calculate the probabilities for each unique value of XX. Since the die is fair and each outcome has an equal probability of 18\frac{1}{8}, we just need to count how many times each value of XX occurs:

  • X=1X = -1 occurs 1 time (OOO)
  • X=3X = 3 occurs 3 times (OOE, OEO, EOO)
  • X=7X = 7 occurs 3 times (OEE, EOE, EEO)
  • X=11X = 11 occurs 1 time (EEE)

So the probability distribution of XX is:

\text{Value } x \text{ of } X & P(X = x) \\ \hline -1 & \frac{1}{8} \\ 3 & \frac{3}{8} \\ 7 & \frac{3}{8} \\ 11 & \frac{1}{8} \\ \end{array}$$ ### Final Answer: The probability distribution of $$X$$ is: - $$P(X = -1) = \frac{1}{8}$$ - $$P(X = 3) = \frac{3}{8}$$ - $$P(X = 7) = \frac{3}{8}$$ - $$P(X = 11) = \frac{1}{8}$$ --- Do you want details or have any questions? Here are some related questions you might consider: 1. What is the expected value of $$X$$? 2. How would the distribution change if the die had more sides? 3. What is the variance of the random variable $$X$$? 4. How would the probabilities change if the die were biased? 5. Can you generalize this problem to $$n$$ rolls? 6. How would the outcome change if we counted odd numbers instead? 7. How does this relate to binomial distributions? 8. Can we visualize this probability distribution? **Tip:** When dealing with probability distributions, always verify that the sum of all probabilities equals 1.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Random Variables
Probability Distributions

Formulas

Probability distribution calculations

Theorems

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Suitable Grade Level

Advanced High School