Two fair dice are thrown simultaneously. Find the probability that the total is 7, the total is a prime number, exactly one of the scores is a 6, the total is at least 8, at least... Two fair dice are thrown simultaneously. Find the probability that the total is 7, the total is a prime number, exactly one of the scores is a 6, the total is at least 8, at least one of the scores is a 6, the two scores are the same, and the difference between the scores is an odd number.

Understand the Problem

The question is asking to find the probabilities of various outcomes when rolling two fair dice. Each option presents a different scenario regarding the sums and values of the dice, indicating a need for probability calculations based on combinations of dice rolls.

Answer

a) $\frac{1}{6}$ b) $\frac{5}{12}$ c) $\frac{5}{12}$ d) $\frac{11}{36}$ e) $\frac{5}{18}$ f) $\frac{1}{6}$ g) $\frac{1}{2}$

Steps to Solve

Determine Total Outcomes

When rolling two dice, each die has 6 faces. Therefore, the total number of outcomes is: $$ 6 \times 6 = 36 $$

Calculate Probability for Each Scenario

a) The total is 7

The combinations that yield a total of 7 are:

(1, 6)
(2, 5)
(3, 4)
(4, 3)
(5, 2)
(6, 1)

Thus, there are 6 successful outcomes. The probability is: $$ P(7) = \frac{6}{36} = \frac{1}{6} $$

b) The total is at least 8

The combinations for totals 8, 9, 10, 11, and 12 are:

Total 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → 5 outcomes
Total 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes
Total 10: (4, 6), (5, 5), (6, 4) → 3 outcomes
Total 11: (5, 6), (6, 5) → 2 outcomes
Total 12: (6, 6) → 1 outcome

Summing these gives: $$ 5 + 4 + 3 + 2 + 1 = 15 $$ So, the probability is: $$ P(\text{at least } 8) = \frac{15}{36} = \frac{5}{12} $$

c) The total is a prime number

The prime totals are 2, 3, 5, 7, 11. The combinations for these are:

Total 2: (1, 1) → 1 outcome
Total 3: (1, 2), (2, 1) → 2 outcomes
Total 5: (1, 4), (2, 3), (3, 2), (4, 1) → 4 outcomes
Total 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 outcomes
Total 11: (5, 6), (6, 5) → 2 outcomes

Summing these outcomes gives: $$ 1 + 2 + 4 + 6 + 2 = 15 $$ So, the probability is: $$ P(\text{prime}) = \frac{15}{36} = \frac{5}{12} $$

d) At least one of the scores is a 6

To find this, we can use the complement rule:

Total outcomes with no 6's: $(1, 1)$, $(1, 2)$, ..., $(5, 5)$ (5 outcomes for each die): $$ 5 \times 5 = 25 $$
Thus, the success outcomes: $$ 36 - 25 = 11 $$ So, the probability is: $$ P(\text{at least one } 6) = \frac{11}{36} $$

e) Exactly one of the scores is a 6

These combinations are:

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)
(1, 6), (2, 6), (3, 6), (4, 6), (5, 6)

Counting gives: $$ 5 + 5 = 10 $$ So, the probability is: $$ P(\text{exactly one } 6) = \frac{10}{36} = \frac{5}{18} $$

f) The two scores are the same

The pairs are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), which gives 6 outcomes. The probability is: $$ P(\text{same}) = \frac{6}{36} = \frac{1}{6} $$

g) The difference between the scores is an odd number

The pairs with an odd difference are those where one die shows an odd number, and the other shows an even number or vice versa (1,2), (1,4), etc.

Count the combinations:

For each odd number (1, 3, 5), you can pair it with (2, 4, 6) → 3 odd × 3 even = 9 outcomes.
Similarly for each even number (2, 4, 6), you can pair it with (1, 3, 5) → Another 9 outcomes.

Thus, total outcomes with an odd difference: $$ 9 + 9 = 18 $$ So, the probability is: $$ P(\text{odd difference}) = \frac{18}{36} = \frac{1}{2} $$

a) $P(\text{total is 7}) = \frac{1}{6}$

b) $P(\text{total is at least 8}) = \frac{5}{12}$

c) $P(\text{total is prime}) = \frac{5}{12}$

d) $P(\text{at least one score is 6}) = \frac{11}{36}$

e) $P(\text{exactly one score is 6}) = \frac{5}{18}$

f) $P(\text{both scores are the same}) = \frac{1}{6}$

g) $P(\text{difference is odd}) = \frac{1}{2}$

More Information

When rolling two dice, the combinations can lead to interesting patterns, such as pairs and sums. Recognizing these patterns helps in calculating probabilities efficiently.

Tips

Confusing the total number of outcomes: Remember there are always $36$ outcomes when rolling two dice, irrespective of the event.
Forgetting complement probabilities: For events like "at least one," it's often easier to calculate the complementary event.

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