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

Question

What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

A. 1/6
B. 1/4
C. 1/2
D. 21/216
E. 32/216

Richie Weasel · Accepted Answer

I am confident that there is a better way, but I looked at the first few possibilities

throw a 3 - 1 way
throw a 4 - 3 ways
throw a 5 - 6 ways
throw a 6 - 10 ways

and recognized these as triangular numbers (the series could also be identified as add 2, add 3, add 4 ...).

from there it was easy to calculate that there are 21 ways to hit an 8 and 15 ways to hit a 14. (Problems with fair dice and coins produce symmetrical probability distributions, so one can count down from 1 way to throw an 18)

gayathri · Answer

The other way would be to list out each option.
The options for a sum of 8: (6,1,1) has 3 options ie (6,1,1), (1,6,1), (1,1,6); (5,2,1) has 6 options, (4,3,1) has 6 options, (4,2,2) has 3 options, (3,3,2) has 3 options. We have 21 options to get 8.

The options for a sum of 14: (6,4,4) has 3 options, (6,5,3) has 6 options, (6,6,2) has 3 options, (5,5,4) has 3 options. We have 15 options to get 14.

Total: 21+15= 36/216 = 1/6.

jinino · Answer

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

gmat1obsessed · Answer

Is there any other to solve this question ?

Can you elaborate the answer pls. ?

Temurkhon · Answer

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

frankb20448 · 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:

2:

…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:

3:

…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.

NaeemHasan · Answer

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

nick28 · Answer

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.

NaeemHasan · Answer

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

PadmajaH · Answer

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

mariamelsakka1 · Answer

where did the 216 come from?

sreyansh · Answer

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?

sharkr · Answer

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?

Bunuel · Answer

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.

KrishnakumarKA1 · Answer

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

Snooopy · Answer

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.

Bunuel · Answer

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

dmorgan1 · Answer

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

3 ways

dmorgan1 · Answer

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

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

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