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.