What is the probability of getting a sum of 8 or 14 when

Wow, Richie, this is a fast and easy way to solve it! I will take a note of this method!

Is there any other to solve this question ?

Can you elaborate the answer pls. ?

I missed some options for 8 as a sum. Idea is that options containing three different number have 3!=6 possible events, options containing 2 identical and 1 other have 3!/2!*1!=3 possible events. By multiplying and summing we get the answer

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.

Hi,
how do you throw 3 in 1 way? I have not understood these points. Will you explain these?

You can throw 3 dices simultaneously to get sum as 3 in only one way - i.e. each dice should have 1 . Hence , you throw 3 in 1 way. And similarly for other numbers.

Now, got that. Can you describe about the triangular system as mentioned in the first reply?

How 15 ways to hit a 14 as per this logic?

where did the 216 come from?

When you say that these are triangular numbers, the series is as follows:

Total number - No. of ways
3 1
4 3
5 6
6 10
7 15
8 21
9 28
10 36
11 36
12 28
13 21
14 15
15 10
16 6
17 3
18 1

Adding them all, total no. of ways : 240

But for 3 throws of a dice, total no. of ways = 6x6x6 = 216

How do you account for the discrepancy?

If we're rolling 3 dice simultaneously, why is it that we're counting, for example, (6,1,1) as 3 ways?
Is it assumed that 3 dice are distinguishable?

Yes. Consider the die to be red, blue and green. Then 6, 1, 1 can occur in 3 ways:

red - blue - green
6 - 1 - 1
1 - 6 - 1
1 - 1 - 6.

Possible outcomes (1,1,6): 3 ways, (1,2,5): 6 ways, (1,3,4): 6 ways, (2,2,4): 3 ways, (2,3,3): 3 ways, (4,4,6): 3 ways, (4,5,5): 3 ways, (5,6,3): 6 ways, (6,6,2): 3 ways
Total outcomes =36
Total outcomes = 6*6*6 = 216
Probability = 36/216 = ⅙ . Option A

I dont understand why there are only three ways to roll (1,1,6). There are three difference dices; therefore there is supposed to be 6 ways.

Please read the whole thread: https://gmatclub.com/forum/what-is-the- ... l#p1799554

1, 1, 6
1, 6, 1
6, 1, 1

3 ways

1 Dice: Probabilities of sums

1: 1/6
2: 1/6
3: 1/6
4: 1/6
5: 1/6
6: 1/6

2 Die: Probabilities of sums

2: 1/36
3: 2/36
4: 3/36
5: 4/36
6: 5/36
7: 6/36
8: 5/36
9: 4/36
10: 3/36
11: 2/36
12: 1/36

Probability of 8 with 3 die: 1 on first then 7 on next two, 2 on first and 6 on next two etc...
Probability of 8 with 3 die = (1/6)[(6 + 5 + 4 + 3 + 2 + 1)/36] = 21/216

Probability of 14 with 3 die: 2 on first then 12 on next two, 3 on first and 11 on next two etc...
Probability of 14 with 3 die = (1/6)[(5 + 4 + 3 + 2 + 1)/36] = 15/216

Probability of 8 OR 14 with 3 die = 21/216 + 15/216 = 36/216 = 1/6