Math Problem Statement

An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is tht, then =Ntht2. Suppose that the random variable X is defined in terms of N as follows: =X−N2−N4. The values of X are given in the table below. Outcome tth hth thh tht hht htt ttt hhh Value of X −2 −4 −4 −2 −4 −2 2 −4 Calculate the probabilities P=Xx 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 P=Xx

Solution

To solve this problem, we need to determine the probabilities for each value of the random variable XX. The probability distribution depends on the frequency of occurrence of each value of XX.

Step-by-Step Solution:

1. List the possible outcomes of tossing the coin three times:

The outcomes of 3 coin tosses are combinations of heads (h) and tails (t). There are 23=82^3 = 8 possible outcomes:

Outcomes={tth,hth,thh,tht,hht,htt,ttt,hhh}\text{Outcomes} = \{ tth, hth, thh, tht, hht, htt, ttt, hhh \}

2. Identify the corresponding values of XX:

We are given the values of XX for each outcome:

\text{Outcome} & : \quad tth, hth, thh, tht, hht, htt, ttt, hhh \\ X & : \quad -2, -4, -4, -2, -4, -2, 2, -4 \end{aligned}$$ #### 3. **Determine the probabilities $$P(X = x)$$:** We now calculate the probability of each value of $$X$$ based on how often each value occurs in the list of outcomes. - The total number of outcomes is $$8$$. - The probability of each outcome is $$\frac{1}{8}$$ because the coin is fair and all outcomes are equally likely. Now, we can count the occurrences of each value of $$X$$ and calculate the probabilities. #### 4. **Count occurrences for each value of $$X$$:** - $$X = -2$$ appears for outcomes: $$tth, tht, htt$$. This happens 3 times, so $$P(X = -2) = \frac{3}{8}$$. - $$X = -4$$ appears for outcomes: $$hth, thh, hht, hhh$$. This happens 4 times, so $$P(X = -4) = \frac{4}{8} = \frac{1}{2}$$. - $$X = 2$$ appears for the outcome $$ttt$$. This happens 1 time, so $$P(X = 2) = \frac{1}{8}$$. #### 5. **Create the table for the probability distribution of $$X$$:** The table of values of $$X$$ and their corresponding probabilities $$P(X = x)$$ is: | Value $$x$$ of $$X$$ | $$-4$$ | $$-2$$ | $$2$$ | |--------------------|-------|-------|-----| | Probability $$P(X = x)$$ | $$\frac{1}{2}$$ | $$\frac{3}{8}$$ | $$\frac{1}{8}$$ | #### Final Answer: The probabilities of $$X = x$$ are: - $$P(X = -4) = \frac{1}{2}$$ - $$P(X = -2) = \frac{3}{8}$$ - $$P(X = 2) = \frac{1}{8}$$ Would you like any further details or clarifications on the calculation? Here are 5 related questions that may expand your understanding: 1. How would the probabilities change if the coin were biased? 2. What is the expected value of $$X$$ for this distribution? 3. How do we calculate the variance of $$X$$ in this case? 4. What if the coin is tossed 4 times instead? How would the random variable $$N$$ and $$X$$ be adjusted? 5. Can you explain how the law of large numbers would apply to repeated experiments of this coin toss? **Tip:** The sum of all probabilities in any probability distribution should always equal 1, which is a good way to check your work.

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Math Problem Analysis

Mathematical Concepts

Probability
Random Variables
Discrete Probability Distribution

Formulas

Probability P(X = x) = Frequency of X / Total Outcomes
X = -N^2 - N^4

Theorems

Law of Total Probability
Concept of Expected Value

Suitable Grade Level

Grades 9-12