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An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.

Compute the probability of each of the following events:

Event [tex]$A$[/tex]: The sum is greater than 9.
Event [tex]$B$[/tex]: The sum is divisible by 3 or 5 (or both).

Round your answers to two decimal places.

(a) [tex]$P(A)=$[/tex] [tex]$\square$[/tex]
(b) [tex]$P(B)=$[/tex] [tex]$\square$[/tex]

Sagot :

GinnyAnsweridn
Let's start by understanding the problem: We have two ordinary (fair) dice, and we roll both dice and sum the resulting numbers. We need to compute the probabilities for two specific events: Event A and Event B.

### Step-by-Step Solution:

1. Determine the Total Number of Possible Outcomes:
Each die has 6 faces, so when two dice are rolled, the total number of possible outcomes is:

[tex]\[ \text{Total possible outcomes} = 6 \times 6 = 36 \][/tex]

2. Event A: The sum is greater than 9

To find the probability of Event A, we need to count the number of outcomes where the sum of the two dice is greater than 9.

- Possible sums greater than 9 are: 10, 11, and 12.
- List the combinations for each of these sums:
- Sum = 10: (4,6), (5,5), (6,4)
- Sum = 11: (5,6), (6,5)
- Sum = 12: (6,6)

- Total number of outcomes for the sum greater than 9 is:
[tex]\[ \text{Count of A} = 3 (\text{for sum 10}) + 2 (\text{sum 11}) + 1 (\text{sum 12}) = 6 \][/tex]

- Therefore, the probability of Event A is:
[tex]\[ P(A) = \frac{\text{Count of A}}{\text{Total possible outcomes}} = \frac{6}{36} = 0.17 \][/tex]

3. Event B: The sum is divisible by 3 or 5 (or both)

To find the probability of Event B, we need to count the number of outcomes where the sum of the two dice is divisible by 3 or 5 (or both).

- Possible sums divisible by 3 or 5 are: 3, 5, 6, 9, 10, 12.
- List the combinations for each of these sums:
- Sum = 3: (1,2), (2,1)
- Sum = 5: (1,4), (2,3), (3,2), (4,1)
- Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1)
- Sum = 9: (3,6), (4,5), (5,4), (6,3)
- Sum = 10: (4,6), (5,5), (6,4)
- Sum = 12: (6,6)

- Total number of outcomes for sums divisible by 3 or 5 is:
[tex]\[ \text{Count of B} = 2 (\text{for sum 3}) + 4 (\text{sum 5}) + 5 (\text{sum 6}) + 4 (\text{sum 9}) + 3 (\text{sum 10}) + 1 (\text{sum 12}) = 19 \][/tex]

- Therefore, the probability of Event B is:
[tex]\[ P(B) = \frac{\text{Count of B}}{\text{Total possible outcomes}} = \frac{19}{36} = 0.53 \][/tex]

### Final Answers:

(a) The probability of Event A:
[tex]\[ P(A) = 0.17 \][/tex]

(b) The probability of Event B:
[tex]\[ P(B) = 0.53 \][/tex]

Rounded to two decimal places, the probabilities are 0.17 for Event A and 0.53 for Event B.
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