Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 (ii) six appears on at least one die

Explanation

To find the probability distribution of the number of successes in two tosses of a die, we will analyze two cases as defined in the question:

**Case (i): Success is defined as rolling a number greater than 4.**

1. **Define the Random Variable:**
Let \( X \) be the number of successes (i.e., the number of times a number greater than 4 appears in two tosses).

2. **Possible Values of \( X \):**
The possible values of \( X \) can be:
- \( X = 0 \): No numbers greater than 4.
- \( X = 1 \): One number greater than 4.
- \( X = 2 \): Both numbers greater than 4.

3. **Sample Space:**
The total sample space when tossing two dice is \( 6 \times 6 = 36 \) outcomes.

4. **Calculate Probabilities:**
- **For \( X = 0 \)**: The outcomes where no number is greater than 4 are (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4). There are 16 outcomes.
\[ P(X = 0) = \frac{16}{36} = \frac{4}{9} \]

- **For \( X = 1 \)**: The outcomes where one number is greater than 4 are (5,1), (5,2), (5,3), (5,4), (1,5), (2,5), (3,5), (4,5), (6,1), (6,2), (6,3), (6,4). There are also 16 outcomes.
\[ P(X = 1) = \frac{16}{36} = \frac{4}{9} \]

- **For \( X = 2 \)**: The outcomes where both numbers are greater than 4 are (5,5), (5,6), (6,5), (6,6). There are 4 outcomes.
\[ P(X = 2) = \frac{4}{36} = \frac{1}{9} \]

5. **Probability Distribution Table:**
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{4}{9} \\ 1 & \frac{4}{9} \\ 2 & \frac{1}{9} \\ \hline \end{array} \]

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**Case (ii): Success is defined as 6 appearing on at least one die.**

1. **Define the Random Variable:**
Let \( Y \) be the number of successes (i.e., the number of times 6 appears in two tosses).

2. **Possible Values of \( Y \):**
The possible values of \( Y \) can be:
- \( Y = 0 \): 6 does not appear on either die.
- \( Y = 1 \): 6 appears on at least one die.

3. **Calculate Probabilities:**
- **For \( Y = 0 \)**: The outcomes where 6 does not appear at all are all combinations of (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5). There are 25 outcomes.
\[ P(Y = 0) = \frac{25}{36} \]

- **For \( Y = 1 \)**: The outcomes where 6 appears at least once are all combinations that include at least one 6. There are 11 outcomes.
\[ P(Y = 1) = \frac{11}{36} \]

4. **Probability Distribution Table:**
\[ \begin{array}{|c|c|} \hline Y & P(Y) \\ \hline 0 & \frac{25}{36} \\ 1 & \frac{11}{36} \\ \hline \end{array} \]

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