To construct the probability distribution of getting heads in a three-coin toss, we need to list all possible outcomes, count the number of heads in each outcome, and then find the probabilities for each possible number of heads.

Step 1: List all possible outcomes.

When tossing three coins, each coin can either be heads (H) or tails (T). The total number of possible outcomes is $2^3 = 8$. Here are the outcomes:

• HHH (3 heads)
• HHT (2 heads)
• HTH (2 heads)
• HTT (1 head)
• THH (2 heads)
• THT (1 head)
• TTH (1 head)
• TTT (0 heads)

Step 2: Count the number of heads for each outcome.

• 3 heads: HHH
• 2 heads: HHT, HTH, THH
• 1 head: HTT, THT, TTH
• 0 heads: TTT

Step 3: Calculate the probabilities.

The probability of each individual outcome is $\frac{1}{8}$, since all outcomes are equally likely. Now, we group the outcomes by the number of heads and calculate the probability for each:

• 0 heads: 1 outcome (TTT) → Probability $P(X = 0) = \frac{1}{8}$
• 1 head: 3 outcomes (HTT, THT, TTH) → Probability $P(X = 1) = \frac{3}{8}$
• 2 heads: 3 outcomes (HHT, HTH, THH) → Probability $P(X = 2) = \frac{3}{8}$
• 3 heads: 1 outcome (HHH) → Probability $P(X = 3) = \frac{1}{8}$

Step 4: Probability Distribution Table

Number of Heads (X)	Probability $P(X)$
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

This table represents the probability distribution of getting heads in a three-coin toss.

Would you like further details or have any questions on this?

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Here are 5 related questions to expand on this:

1. What is the expected number of heads in a three-coin toss?
2. How would the distribution change for a four-coin toss?
3. What is the probability of getting at least two heads?
4. How can this distribution be modeled using a binomial distribution?
5. What is the cumulative probability of getting fewer than 2 heads?

Tip: A binomial distribution can be used for any number of coin tosses, with heads or tails representing a success or failure in each trial.