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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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26 Nov 2004, 06:51

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Quote:

"Gayathri" 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

Is there a shorter way to do this than listing each possibility?

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)

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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26 Nov 2004, 22:03

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Richie Weasel wrote:

Quote:

"Gayathri" 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

Is there a shorter way to do this than listing each possibility?

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)

[more than 2 minutes - less than 3 minutes]

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

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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28 Nov 2004, 05:05

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gmat+obsessed wrote:

Is there any other to solve this question ?

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.

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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12 Sep 2014, 05:45

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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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22 Oct 2014, 04:17

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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

What is the probability of getting a sum of 8 or 14 when [#permalink]

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17 Nov 2014, 13:03

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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 \(\frac{1}{6}\).

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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04 Dec 2015, 02:27

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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26 Aug 2016, 10:10

Richie Weasel wrote:

Quote:

"Gayathri" 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

Is there a shorter way to do this than listing each possibility?

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)

[more than 2 minutes - less than 3 minutes]

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

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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27 Aug 2016, 20:54

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 wrote:

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

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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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30 Aug 2016, 07:08

nick28 wrote:

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 wrote:

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

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

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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14 Sep 2016, 22:46

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Richie Weasel wrote:

Quote:

"Gayathri" 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

Is there a shorter way to do this than listing each possibility?

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)

Re: What is the probability of getting a sum of 8 or 14 when [#permalink]

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30 Sep 2016, 13:48

Richie Weasel wrote:

Quote:

"Gayathri" 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

Is there a shorter way to do this than listing each possibility?

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)

[more than 2 minutes - less than 3 minutes]

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

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