Questions: 6. Let S represent the sum of the rolls of a pair of fair six-sided dice. Use probability notation and inequalities to present the probability of each of the following events (as fractions). a. The sum is exactly 6. 5/36 b. The sum is more than 12. c. The sum is no more than 10. 7. Let H be the number of times a fair coin shows heads when tossed thrice. Use probability notation and inequalities to present the probability of the following events (as fractions). a. The coin lands on heads exactly once. b. The coin lands on heads more than once. c. The coin lands on heads at most once.

Solution Steps

To solve these problems, we need to calculate probabilities based on the outcomes of rolling dice and tossing coins. For the dice, we consider all possible outcomes of rolling two six-sided dice and count the favorable outcomes for each event. For the coin toss, we consider all possible outcomes of tossing a coin three times and count the favorable outcomes for each event.

1. Sum is exactly 6: Count the number of outcomes where the sum of the dice is 6 and divide by the total number of outcomes (36).
2. Sum is more than 12: Since the maximum sum with two dice is 12, this probability is 0.
3. Sum is no more than 10: Count the number of outcomes where the sum is 10 or less and divide by the total number of outcomes (36).

1. Heads exactly once: Count the number of outcomes with exactly one head in three tosses and divide by the total number of outcomes (8).
2. Heads more than once: Count the number of outcomes with more than one head and divide by the total number of outcomes (8).
3. Heads at most once: Count the number of outcomes with at most one head and divide by the total number of outcomes (8).

To find the probability that the sum of two six-sided dice is exactly 6, we count the favorable outcomes: \((1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\). There are 5 such outcomes.

The total number of possible outcomes when rolling two dice is 36. Thus, the probability is:

\[ P(S = 6) = \frac{5}{36} \]

The maximum sum possible with two six-sided dice is 12. Therefore, the probability of the sum being more than 12 is 0.

\[ P(S > 12) = 0 \]

To find the probability that the sum is no more than 10, we count the outcomes where the sum is 10 or less. There are 33 such outcomes.

Thus, the probability is:

\[ P(S \leq 10) = \frac{33}{36} = \frac{11}{12} \]

For three coin tosses, the outcomes with exactly one head are: \((H, T, T), (T, H, T), (T, T, H)\). There are 3 such outcomes.

The total number of possible outcomes is 8. Thus, the probability is:

\[ P(H = 1) = \frac{3}{8} \]

The outcomes with more than one head are: \((H, H, T), (H, T, H), (T, H, H), (H, H, H)\). There are 4 such outcomes.

Thus, the probability is:

\[ P(H > 1) = \frac{4}{8} = \frac{1}{2} \]

The outcomes with at most one head are: \((T, T, T), (H, T, T), (T, H, T), (T, T, H)\). There are 4 such outcomes.

Thus, the probability is:

\[ P(H \leq 1) = \frac{4}{8} = \frac{1}{2} \]

Final Answer