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kolbaska11 [484]
3 years ago
9

22 points + brainliest! A fair die with sides labeled 1 through 6 is rolled two times. The values of the two rolls are added tog

ether. The sum is recorded as the outcome of a single trial of a random experiment. Compute the probability that the sum is 9.

Mathematics
2 answers:
murzikaleks [220]3 years ago
8 0

Answer:

P(9) = 1/9

Step-by-step explanation:

From the contingency table, we see that 9 appears 4 times out of the 36 possible outcomes, therefore the probability of having a sum of 9 is

P(9) = 4/36 = 1/9

ki77a [65]3 years ago
3 0

If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.

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Answer: - 0.28


Explanation:

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That is: ∑ of prbability of event i × value of event i.

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2) Sample space:

Sum              Points awarded
    
1+ 1 = 2              +6

1 + 2 = 3             +2

1 + 3 = 4             -1

1 + 4 = 5             -1

1 + 5 = 6             -1

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2 + 1 = 3             +2

2 + 2 = 4             -1

2 + 3 = 5             -1

2 + 4 = 6             -1

2 + 5 = 7             -1

2 + 6 = 8             -1

3 + 1 = 4             -1

3 + 2 = 5             -1

3 + 3 = 6             -1

3 + 4 = 7             -1

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3 + 6 = 9             -1

4 + 1 = 5             -1

4 + 2 = 6             -1

4 + 3 = 7             -1

4 + 4 = 8            -1

4 + 5 = 9            -1

4 + 6 = 10          -1

5 + 1 = 6            -1

5 + 2 = 7            -1

5 + 3 = 8            -1

5 + 4 = 9            -1

5 + 5 = 10          -1

5 + 6 = 11          +2

6 + 1 = 7            -1

6 + 2 = 8            -1

6 + 3 = 9            -1

6 + 4 = 10          -1

6 + 5 = 11          +2

6 + 6 = 12          +6


2) Probabilities

From that, there is:
- 2/36 probabilities to earn + 6 points.
- 4/36 probabilites to earn + 2 points
- the rest, 30/36 probabilities to earn - 1 points


3) Expected value = (2/36)(+6) + (4/36) (+2) + (30/36) (-1) = - 0.28
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