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 htt, then =Nhtt2. Suppose that the random variable X is defined in terms of N as follows: =X−N−N22. The values of X are given in the table below.
Outcome	hth	ttt	tth	thh	hhh	tht	hht	htt
Value of X	−2	−8	−4	−2	−2	−4	−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 calculate the probabilities \( P(X = x) \), we first need to understand the distribution of the values of \( X \) and how often each value appears in the set of all possible outcomes of tossing a fair coin three times.

### Step 1: List the outcomes and their corresponding values of \( X \)

From the problem statement, we have:

| Outcome | Value of \( X \) |
|---------|------------------|
| hth     | -2               |
| ttt     | -8               |
| tth     | -4               |
| thh     | -2               |
| hhh     | -2               |
| tht     | -4               |
| hht     | -2               |
| htt     | -4               |

### Step 2: Calculate the probabilities

The probability of each outcome in a fair coin toss is the same. Since there are \( 2^3 = 8 \) possible outcomes, the probability of any specific outcome is:

\[
P(\text{any specific outcome}) = \frac{1}{8}
\]

Now, we group the outcomes based on their corresponding value of \( X \).

- For \( X = -2 \), the outcomes are: hth, thh, hhh, hht. There are 4 outcomes, so the probability is:
  \[
  P(X = -2) = \frac{4}{8} = \frac{1}{2}
  \]

- For \( X = -4 \), the outcomes are: tth, tht, htt. There are 3 outcomes, so the probability is:
  \[
  P(X = -4) = \frac{3}{8}
  \]

- For \( X = -8 \), the outcome is: ttt. There is 1 outcome, so the probability is:
  \[
  P(X = -8) = \frac{1}{8}
  \]

### Step 3: Fill in the table

Now we can fill in the table with the values of \( X \) and the corresponding probabilities:

| Value \( x \) of \( X \) | -8  | -4  | -2  |
|--------------------------|-----|-----|-----|
| Probability \( P(X = x) \)| 1/8 | 3/8 | 4/8 |

Thus, the probability distribution of \( X \) is:

- \( P(X = -2) = \frac{4}{8} = \frac{1}{2} \)
- \( P(X = -4) = \frac{3}{8} \)
- \( P(X = -8) = \frac{1}{8} \)

### Would you like details on how the grouping was done, or have any questions about this process?

### 5 Relative Questions:
1. How would the probabilities change if the coin were biased?
2. Can you calculate the expected value of \( X \)?
3. What is the variance of \( X \) for this probability distribution?
4. How would the results differ if the coin were tossed 4 times instead of 3?
5. What is the probability that \( X \leq -4 \)?

### Tip:
When working with probability distributions, always ensure that the sum of all probabilities equals 1.

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 htt, then N(htt) = 2. Suppose that the random variable X is defined in terms of N as follows: X = −N − N^2. The values of X are given in the table below. Outcome: hth, ttt, tth, thh, hhh, tht, hht, htt; Value of X: −2, −8, −4, −2, −2, −4, −2, −4. Calculate the probabilities P(X=x) of the probability distribution of X.

Calculate the probability distribution of a random variable X defined as X = −N − N^2, where N counts the number of tails in 3 tosses of a fair coin. This detailed solution covers key concepts in probability theory and is suitable for Grades 9-12.

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