1.8k views
2 votes
A pair of 6 sided dice are tossed. What is the probability that at least one of the dice has a value greater than or equal to 2 ? Answer = In the rolling of two fair dice calculate the following: P (Sum of the two dice is 8)= P (Sum of the two dice is 6)= P (Sum of the two dice is not 7)= P (Sum of the two dice is 10 or 9)= P( Sum of the two dice is not 4 and not 5)=

User Rhak Kahr
by
7.5k points

1 Answer

3 votes

**Probability that at least one die has a value greater than or equal to 2:**

Since all sides of a fair 6-sided die have values greater than or equal to 2, the probability that at least one die meets this condition is 1 (or 100%).

**Probability of specific sums:**

- P(Sum of the two dice is 8):
There are 5 ways to achieve a sum of 8: (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2). Since each die has 6 sides, there are a total of 6 * 6 = 36 possible outcomes when rolling two dice. Therefore, P(Sum of 8) = 5/36.

- P(Sum of the two dice is 6):
There are 5 ways to get a sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1). So, P(Sum of 6) = 5/36.

- P(Sum of the two dice is not 7):
There are 6 ways to get a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). Since there are a total of 36 possible outcomes, P(Sum is not 7) = 1 - P(Sum of 7) = 1 - 6/36 = 30/36.

- P(Sum of the two dice is 10 or 9):
There are 4 ways to get a sum of 10: (4, 6), (5, 5), (6, 4), and (6, 3). There are also 4 ways to get a sum of 9: (3, 6), (4, 5), (5, 4), and (6, 3). So, P(Sum is 10 or 9) = (4 + 4) / 36 = 8/36.

- P(Sum of the two dice is not 4 and not 5):
There are 2 ways to get a sum of 4: (1, 3) and (3, 1), and 4 ways to get a sum of 5: (1, 4), (2, 3), (3, 2), and (4, 1). So, P(Sum is not 4 and not 5) = 1 - P(Sum of 4) - P(Sum of 5) = 1 - (2/36) - (4/36) = 30/36.
User Mpluse
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.