Two fair six-sided dice are tossed independently. Let M be the maximum of two
tosses. What is the probablity mass function of M.

Let's find the probability mass function (PMF) of \( M \), the maximum of the two tosses of fair six-sided dice.

### Step 1: Understanding the Sample Space
When two six-sided dice are rolled, each die can show any of the numbers \( 1, 2, 3, 4, 5, 6 \). The total number of possible outcomes is \( 6 \times 6 = 36 \).

### Step 2: Defining the Random Variable \( M \)
The random variable \( M \) represents the maximum value obtained from the two dice. \( M \) can take values from 1 to 6, inclusive.

### Step 3: Finding the Probability for Each Value of \( M \)
We will calculate the probability \( P(M = m) \) for each possible value of \( M \):

- **\( P(M = 1) \):**
  - \( M = 1 \) occurs only if both dice show 1.
  - Probability \( P(M = 1) = \frac{1}{36} \).

- **\( P(M = 2) \):**
  - \( M = 2 \) occurs if at least one die shows 2, and the other die shows 1 or 2.
  - Possible outcomes: \( (1,2), (2,1), (2,2) \).
  - Probability \( P(M = 2) = \frac{3}{36} = \frac{1}{12} \).

- **\( P(M = 3) \):**
  - \( M = 3 \) occurs if at least one die shows 3, and the other die shows 1, 2, or 3.
  - Possible outcomes: \( (1,3), (2,3), (3,1), (3,2), (3,3) \).
  - Probability \( P(M = 3) = \frac{5}{36} \).

- **\( P(M = 4) \):**
  - \( M = 4 \) occurs if at least one die shows 4, and the other die shows 1, 2, 3, or 4.
  - Possible outcomes: \( (1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (4,4) \).
  - Probability \( P(M = 4) = \frac{7}{36} \).

- **\( P(M = 5) \):**
  - \( M = 5 \) occurs if at least one die shows 5, and the other die shows 1, 2, 3, 4, or 5.
  - Possible outcomes: \( (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) \).
  - Probability \( P(M = 5) = \frac{9}{36} = \frac{1}{4} \).

- **\( P(M = 6) \):**
  - \( M = 6 \) occurs if at least one die shows 6, and the other die shows any number from 1 to 6.
  - Possible outcomes: \( (1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \).
  - Probability \( P(M = 6) = \frac{11}{36} \).

### Step 4: Summarizing the PMF
The probability mass function \( P(M = m) \) for \( M \) is given by:

\[
P(M = m) =
\begin{cases}
\frac{1}{36} & \text{if } m = 1, \\
\frac{3}{36} = \frac{1}{12} & \text{if } m = 2, \\
\frac{5}{36} & \text{if } m = 3, \\
\frac{7}{36} & \text{if } m = 4, \\
\frac{9}{36} = \frac{1}{4} & \text{if } m = 5, \\
\frac{11}{36} & \text{if } m = 6.
\end{cases}
\]

This PMF gives the probability distribution of the maximum value obtained when two six-sided dice are rolled.

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Would you like more details or have any questions about this? Here are five related questions you might find interesting:

1. How would the PMF change if the dice were not fair?
2. What is the expected value of the maximum \( M \)?
3. What is the cumulative distribution function (CDF) of \( M \)?
4. How does the distribution of \( M \) compare to the minimum of two dice rolls?
5. How would the PMF change if three dice were rolled instead of two?

**Tip:** When calculating probabilities, breaking down the problem step by step and listing all possible outcomes can greatly simplify the process.

Explore the probability mass function (PMF) of the maximum value obtained from two fair six-sided dice rolls. Understand the distribution of probabilities for each possible outcome, ranging from 1 to 6. Suitable for Grades 10-12.

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