Two fair dice are thrown simultaneously. Find the probability that a) the total is 7, b) the total is at least 8, c) the total is a prime number, d) at least one of the scores is a... Two fair dice are thrown simultaneously. Find the probability that a) the total is 7, b) the total is at least 8, c) the total is a prime number, d) at least one of the scores is a 6, e) exactly one of the scores is a 6, f) the two scores are the same, g) the difference between the scores is an odd number.
Understand the Problem
The question is asking for the probabilities associated with different outcomes from rolling two fair dice. Each option presents a distinct probability scenario related to the dice results, such as the total, characteristics of the scores, and their differences.
Answer
a) \( \frac{1}{6} \) b) \( \frac{5}{12} \) c) \( \frac{7}{18} \) d) \( \frac{11}{36} \) e) \( \frac{5}{18} \) f) \( \frac{1}{6} \) g) \( \frac{1}{2} \)
Answer for screen readers
a) (P(\text{total is 7}) = \frac{1}{6})
b) (P(\text{total at least 8}) = \frac{5}{12})
c) (P(\text{total is prime}) = \frac{7}{18})
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{two scores same}) = \frac{1}{6})
g) (P(\text{difference is odd}) = \frac{1}{2})
Steps to Solve
- Determine the total outcomes
When rolling two fair dice, each die has 6 faces. Therefore, the total number of outcomes is: $$ 6 \times 6 = 36 $$
- Calculate the probability for total equals 7
The combinations to get a total of 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). This gives us 6 favorable outcomes: $$ P(\text{total is 7}) = \frac{6}{36} = \frac{1}{6} $$
- Calculate the probability for total at least 8
The combinations for totals of 8, 9, 10, 11, and 12 are:
- Total of 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- Total of 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- Total of 10: (4,6), (5,5), (6,4) → 3 outcomes
- Total of 11: (5,6), (6,5) → 2 outcomes
- Total of 12: (6,6) → 1 outcome
Adding these gives 5 + 4 + 3 + 2 + 1 = 15 favorable outcomes: $$ P(\text{total at least 8}) = \frac{15}{36} = \frac{5}{12} $$
- Calculate the probability for total being a prime number
The prime totals possible with two dice are 3, 5, 7, 11. The combinations for these are:
- Total of 3: (1,2), (2,1) → 2 outcomes
- Total of 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
- Total of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
- Total of 11: (5,6), (6,5) → 2 outcomes
Totaling gives us 2 + 4 + 6 + 2 = 14 favorable outcomes: $$ P(\text{total is prime}) = \frac{14}{36} = \frac{7}{18} $$
- Calculate the probability for at least one score being a 6
The combinations where at least one die shows 6:
- Case 1: First die is 6 → (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) → 6 outcomes
- Case 2: Second die is 6 → (1,6), (2,6), (3,6), (4,6), (5,6) → 5 outcomes (note (6,6) counted above)
This gives us 6 + 5 = 11 favorable outcomes: $$ P(\text{at least one score is 6}) = \frac{11}{36} $$
- Calculate the probability for exactly one score being a 6
This can occur in two scenarios:
- (6, x) where x is not 6 → 5 outcomes: (6,1), (6,2), (6,3), (6,4), (6,5)
- (x, 6) where x is not 6 → 5 outcomes: (1,6), (2,6), (3,6), (4,6), (5,6)
Adding these gives us 5 + 5 = 10 favorable outcomes: $$ P(\text{exactly one score is 6}) = \frac{10}{36} = \frac{5}{18} $$
- Calculate the probability for the two scores being the same
This is the outcome where both scores are equal. The pairs are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes: $$ P(\text{two scores same}) = \frac{6}{36} = \frac{1}{6} $$
- Calculate the probability for the difference being odd
The difference between the two scores is odd when one score is odd and the other is even. Since there are 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6) on a die:
- There are 3 (odd) x 3 (even) = 9 combinations for odd-even pairs.
- There are also 3 (even) x 3 (odd) = 9 combinations for even-odd pairs.
So, total favorable outcomes = 9 + 9 = 18: $$ P(\text{difference is odd}) = \frac{18}{36} = \frac{1}{2} $$
a) (P(\text{total is 7}) = \frac{1}{6})
b) (P(\text{total at least 8}) = \frac{5}{12})
c) (P(\text{total is prime}) = \frac{7}{18})
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{two scores same}) = \frac{1}{6})
g) (P(\text{difference is odd}) = \frac{1}{2})
More Information
Rolling two dice creates a wide variety of combinations, yielding interesting probabilities for various outcomes. The total outcomes (36) allow us to count favorable combinations for different conditions. This exercise reinforces foundational concepts in probability and combinatorics.
Tips
- Counting outcomes incorrectly: Be careful to list all combinations clearly to avoid missing cases.
- Misunderstanding conditions, such as confusing "at least one" with "exactly one."
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