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Damon rolls three six-sided dice. What is the probability
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09 Dec 2017, 18:26
09 Dec 2017, 18:26
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Question Stats:
68% (01:15) correct
31% (01:55) wrong based on 47 sessions
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Damon rolls three six-sided dice. What is the probability that his total will be greater than 16 ?
Drill 3
Question: 4
Page: 528
Question: 4
Page: 528
Show: :: OA
\(\frac{1}{54}\)
Re: Damon rolls three six-sided dice. What is the probability
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29 Dec 2017, 14:18
29 Dec 2017, 14:18
2
Expert Reply
Explanation
First, figure out how many different results that Damon can get: Each die has 6 sides, so the total number of possible outcomes is 6 × 6 × 6 = 216.
Now count out how many of those outcomes total more than 16. There are 3 ways to roll a 17—5, 6, and 6, 6, 5, and 6, and 6, 6, and 5—and 1 way to roll an 18—three 6’s. The probability is thus \(\frac{4}{216}\), which reduces to \(\frac{1}{54}\).
First, figure out how many different results that Damon can get: Each die has 6 sides, so the total number of possible outcomes is 6 × 6 × 6 = 216.
Now count out how many of those outcomes total more than 16. There are 3 ways to roll a 17—5, 6, and 6, 6, 5, and 6, and 6, 6, and 5—and 1 way to roll an 18—three 6’s. The probability is thus \(\frac{4}{216}\), which reduces to \(\frac{1}{54}\).
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Re: Damon rolls three six-sided dice. What is the probability
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14 May 2019, 08:20
14 May 2019, 08:20
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sandy wrote:
Damon rolls three six-sided dice. What is the probability that his total will be greater than 16 ?
Drill 3
Question: 4
Page: 528
Question: 4
Page: 528
Show: :: OA
\(\frac{1}{54}\)
There are FOUR ways to get a total that's greater than 16:
case i) 1st die is 5, 2nd die is 6 and 3rd die is 6 (aka 5-6-6)
case ii) 1st die is 6, 2nd die is 5 and 3rd die is 6 (aka 6-5-6)
case iii) 1st die is 6, 2nd die is 6 and 3rd die is 5 (aka 6-6-5)
case iv) 1st die is 6, 2nd die is 6 and 3rd die is 6 (aka 6-6-6)
case i:
P(1st die is 5 and 2nd die is 6 and 3rd die is 6) = P(1st die is 5) x P(2nd die is 6) x P(3rd die is 6)
= 1/6 x 1/6 x 1/6
= 1/216
We can quickly see that P(case ii) = 1/216, P(case iii) = 1/216, and O(case iv) = 1/216
So, P(sum greater than 16) = P(case i OR case ii OR case iii OR case iv)
= P(case i) + P(case ii) + P(case iii) + P(case iv)
= 1/216 + 1/216 + 1/216 + 1/216
= 4/216
ASIDE: On the GRE, we need not simplify our fractions.
So, entering 4/216 as our answer would be correct.
In fact, entering 40/2160 as our answer would also be correct.
Or, if we're so inclined, we can simplify 4/216 to get 1/54, which would also be correct.
Answer: 4/216 (aka 1/54)
Cheers,
Brent
