Problem: Coin Toss Probability Distribution

You are tossing a fair coin three times, and the random variable $Y$ represents the number of tails observed in the three tosses. We are asked to construct a probability distribution for $Y$ and determine whether the distribution is discrete.

Step 1: Sample Space

When a fair coin is tossed three times, each toss has two possible outcomes: heads (H) or tails (T). The sample space $S$ of all possible outcomes for three tosses is:

$S = \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$

Thus, there are 8 possible outcomes in total.

Step 2: Counting the Number of Tails (Y)

For each outcome in the sample space, count the number of tails:

• $HHH$: 0 tails
• $HHT$: 1 tail
• $HTH$: 1 tail
• $HTT$: 2 tails
• $THH$: 1 tail
• $THT$: 2 tails
• $TTH$: 2 tails
• $TTT$: 3 tails

So, the possible values of $Y$ (the number of tails) are 0, 1, 2, and 3.

Step 3: Probability Distribution

Now, we need to calculate the probability for each value of $Y$. We will count the occurrences of each number of tails in the sample space and divide by the total number of outcomes (which is 8).

• $P(Y = 0)$: There is 1 outcome with 0 tails: $HHH$. So, $P(Y = 0) = \frac{1}{8}$.
• $P(Y = 1)$: There are 3 outcomes with 1 tail: $HHT, HTH, THH$. So, $P(Y = 1) = \frac{3}{8}$.
• $P(Y = 2)$: There are 3 outcomes with 2 tails: $HTT, THT, TTH$. So, $P(Y = 2) = \frac{3}{8}$.
• $P(Y = 3)$: There is 1 outcome with 3 tails: $TTT$. So, $P(Y = 3) = \frac{1}{8}$.

Thus, the probability distribution for $Y$ is:

\hline Y & P(Y) \\ \hline 0 & \frac{1}{8} \\ 1 & \frac{3}{8} \\ 2 & \frac{3}{8} \\ 3 & \frac{1}{8} \\ \hline \end{array}$$ ### Step 4: Discreteness of the Distribution This is a **discrete probability distribution**, as the random variable $$Y$$ takes on a finite set of values (0, 1, 2, 3), and the probabilities are assigned to each specific value. ### Conclusion - The probability distribution is **discrete**. - The probability distribution is as follows: $$P(Y = 0) = \frac{1}{8}, \quad P(Y = 1) = \frac{3}{8}, \quad P(Y = 2) = \frac{3}{8}, \quad P(Y = 3) = \frac{1}{8}$$ --- Would you like further details or explanations? Here are some related questions: 1. What is the expected value (mean) of the number of tails? 2. How do you calculate the variance of the number of tails? 3. How would the probability distribution change if the coin was biased? 4. What is the probability of getting at least 2 tails? 5. How can you use the binomial distribution formula to calculate this? **Tip:** The sum of all probabilities in a probability distribution must equal 1. You can check your work by adding the probabilities for all possible outcomes.