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

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

### 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 \(2^3 = 8\) possible outcomes:

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

#### 2. **Identify the corresponding values of \(X\):**
We are given the values of \(X\) for each outcome:

\[
\begin{aligned}
\text{Outcome} & : \quad tth, hth, thh, tht, hht, htt, ttt, hhh \\
X & : \quad -2, -4, -4, -2, -4, -2, 2, -4
\end{aligned}
\]

#### 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.

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 N(tht) = 2. Suppose that the random variable X is defined in terms of N as follows: X = −N^2 − N^4. The values of X are given in the table below. Calculate the probabilities P(X = x) of the probability distribution of X.

Solve a probability distribution problem where X is defined based on the number of tails in 3 coin tosses. Learn how to calculate P(X = x) with a step-by-step breakdown, using key concepts of discrete random variables and probability theory. Suitable for Grades 9-12 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation