What is the probability of getting a sum of 8 or 14 when
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17 Nov 2014, 13:03
Here's another way:
You start by listing out each possible triplet for every number on the dice for each of the two numbers we need (notice that for each triplet the probability is \(\frac{1}{6^3}\)):
8:
1:
[1,1,6]
[1,2,5]
[1,3,4]
[1,4,3]
[1,5,2]
[1,6,1]
2:
[2,1,5]
[2,2,4]
[2,3,3]
[2,4,2]
[2,5,1]
…and so on...
Soon thereafter, you realize that, for each number of the dice, there is one triplet less that adds up to 8, than the previous number (i.e. 1-triplets: 6, 2-triplets: 5, 3-triplets: 4, and so on…). Given that \(\frac{1}{6^3}\) is a common factor to all triplets, we get that: \(\frac{1}{6^3}*(6+5+4+3+2+1)\) or \(\frac{1}{6^3}*(21)\).
Then we do a similar process for 14:
1:
(No possible combination adds up to 14)
2:
[2,6,6]
3:
[3,5,6]
[3,6,5]
…and so on…
So you'll notice that a similar thing happens in this case: from 2 on, for each number of the dice, there is one triplet more that adds up to 14, than the previous number (i.e. 2-triplet: 1, 3-triplets: 2, 4-triplets: 3, and so on…). Again, given that \(\frac{1}{6^3}\) is a common factor to all triplets, we get that: \(\frac{1}{6^3}*(1+2+3+4+5)\) or \(\frac{1}{6^3}*(15)\).
Given that we need the probability of getting a sum of 8 OR 14, we add up both of these cases:
\(\frac{1}{6^3}*(21)+\frac{1}{6^3}*(15)\)
We factor out \(\frac{1}{6^3}\), and find that:
\(\frac{1}{6^3}*(36)\), and by recognizing that \(36 = 6^2\), we cross it out with \(6^3\) to find that the probability of getting a sum of 8 or 14 when rolling three fair dice is \(\frac{1}{6}\).
Answer A.