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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 7.
Event [tex]$B$[/tex]: The sum is not divisible by 4 and not divisible by 5.

Round your answers to two decimal places.

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

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

Sagot :

GinnyAnswer
To find the probabilities of the given events when rolling a fair die twice, we will follow a sequence of logical steps:

### Step 1: List all possible outcomes
When you roll a die twice, each roll has 6 possible outcomes. Therefore, the total number of outcomes for rolling the die twice is:
[tex]\[ 6 \times 6 = 36. \][/tex]
The possible sums (2 to 12) and their frequencies are counted as below:
- Sum = 2 : (1,1)
- Sum = 3 : (1,2), (2,1)
- Sum = 4 : (1,3), (2,2), (3,1)
- Sum = 5 : (1,4), (2,3), (3,2), (4,1)
- Sum = 6 : (1,5), (2,4), (3,3), (4,2), (5,1)
- Sum = 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Sum = 8 : (2,6), (3,5), (4,4), (5,3), (6,2)
- Sum = 9 : (3,6), (4,5), (5,4), (6,3)
- Sum = 10 : (4,6), (5,5), (6,4)
- Sum = 11 : (5,6), (6,5)
- Sum = 12 : (6,6)

### Step 2: Compute [tex]$P(A)$[/tex], where Event [tex]$A$[/tex] is that the sum is greater than 7
Firstly, identify which sums are greater than 7: 8, 9, 10, 11, and 12.
- Number of ways to get each sum:
- Sum = 8 : 5 ways
- Sum = 9 : 4 ways
- Sum = 10 : 3 ways
- Sum = 11 : 2 ways
- Sum = 12 : 1 way

Total favorable outcomes for Event [tex]$A$[/tex] = 5 (8) + 4 (9) + 3 (10) + 2 (11) + 1 (12) = 15.

Probability of Event [tex]$A$[/tex]:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for Event } A}{\text{Total number of outcomes}} = \frac{15}{36} \approx 0.42 \][/tex]

Thus:
[tex]\[ P(A) = 0.42 \][/tex]

### Step 3: Compute [tex]$P(B)$[/tex], where Event [tex]$B$[/tex] is that the sum is not divisible by 4 and not divisible by 5
Identify the sums that are not divisible by 4 and not by 5:
- Exclude sums divisible by 4: 4, 8, 12
- Exclude sums divisible by 5: 5, 10

Therefore, sums that are not divisible by 4 or 5: 2, 3, 6, 7, 9, 11

Number of ways to get each sum:
- Sum = 2 : 1 way
- Sum = 3 : 2 ways
- Sum = 6 : 5 ways
- Sum = 7 : 6 ways
- Sum = 9 : 4 ways
- Sum = 11 : 2 ways

Total favorable outcomes for Event [tex]$B$[/tex] = 1 (2) + 2 (3) + 5 (6) + 6 (7) + 4 (9) + 2 (11) = 20.

Probability of Event [tex]$B$[/tex]:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for Event } B}{\text{Total number of outcomes}} = \frac{20}{36} \approx 0.56 \][/tex]

Thus:
[tex]\[ P(B) = 0.56 \][/tex]

### Summary of Results:
(a) \( P(A) = 0.42 \)

(b) [tex]\( P(B) = 0.56 \)[/tex]
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