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If four dice are thrown simultaneously,what is the probability that

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If four dice are thrown simultaneously,what is the probability that [#permalink]

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New post 27 Oct 2010, 06:59
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A
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D
E

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Question Stats:

60% (02:32) correct 40% (02:17) wrong based on 87 sessions

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If four dice are thrown simultaneously,what is the probability that the sum of the number is exactly 20?

A. 35/1296
B. 55/1296
C. 51/1322
D. 61/533
E. 41/1321
[Reveal] Spoiler: OA

Last edited by ankitranjan on 28 Oct 2010, 00:15, edited 1 time in total.

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Re: If four dice are thrown simultaneously,what is the probability that [#permalink]

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New post 27 Oct 2010, 13:59
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ankitranjan wrote:
If four dice are thrown simultaneously,what is the probability that the sum of the number is exactly 20?

1. 35/1296
2. 55/1296
3. 51/1322
4. 61/533
5. 41/1321


Consider KUDOS if u find it as a good question.



We need a sum of 20. This can be achieved in the following ways :
{5,5,5,5} --> 1 way
{5,5,4,6} --> 4!/(2!) ways = 12
{4,4,6,6} --> 4!/(2!2!) ways = 6 ways
{3,5,6,6} --> 4!/2! ways = 12 ways
{2,6,6,6} --> 4!/3! ways = 4 ways

Total ways = 6x6x6x6

So probability = 35/1296 ... Answer (a)
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Kudos [?]: 1212 [0], given: 25

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Re: If four dice are thrown simultaneously,what is the probability that [#permalink]

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New post 28 Oct 2010, 07:37
Ok guys. I Have developed a much much easier way to find the solution and I need many KUDOS for this if you like it:

Dice problem

A nice way to find the sum:

e.g 4 dices are thrown

then the outcome of the sums can be:

4,5,6,………………14……………..,22,23,24

Mid point is 14. The left side of the mid point is mirror image of the right part.

So lets say the probability of getting a 20 is same as the probability of getting a 8

In this case the number of ways to get the sums are:

4: C(3,0) = 1 (start 1 before the # of throws)
5: C(4,1) = 4
6: C(5,2) = 10
7: C(6,3) = 20
8: C(7,4) = 35

So probability of getting a 20 = probability of getting a 8 = 35/6^4 = 35/1296
Similarly for rolloing of 3 dices:

then the outcome of the sums can be:
Mid point is 10.5. The left side of the mid point is mirror image of the right part.

3,4,5,6,………………10.5……………..,15.16.17.18

In this case the number of ways to get the sums are:

3: C(2,0) = 1 (start 1 before the # of throws)
4: C(3,1) = 3
5: C(4,2) = 6
6: C(5,3) = 10
7: C(6,4) = 15
8: C(7,5) = 21

Kudos [?]: 60 [0], given: 0

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Re: If four dice are thrown simultaneously,what is the probability that [#permalink]

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New post 03 May 2017, 12:55
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Hi All,

This question doesn't require any complex math at all; it can be solved with some arithmetic, some 'brute force' and a bit of logic:

To start, each of the 4 dice has 6 possible outcomes (1, 2, 3, 4, 5 and 6), so there are 6^4 = 36^2 = 1296 total outcomes. Thus, the correct answer is either a fraction out of 1296 or a fraction that has been reduced from 1296. This allows you to immediately eliminate answers C, D and E.

From the two remaining answers, there are either 35 or 55 dice combinations that will total 20. Using a bit of 'brute force', we can name the groups:

6662
6653
6644
6554
5555

Each of these groups has a certain number of possible outcomes. For example, 5555 can only occur 1 time, while 6662 can occur 4 times (2666, 6266, 6626 and 6662). At this point, you should note that 2 of the 5 possibilities yield just 1+4 = 5 options, so it's highly likely that the total number of possibilities is relatively small (in this case, 35 and not 55). If you wanted to list out all of the possibilities, then you could (and there would be 35 of them). If you 'time' your work, you could probably complete this step in 1-2 minutes.

Final Answer:
[Reveal] Spoiler:
A


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Re: If four dice are thrown simultaneously,what is the probability that [#permalink]

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New post 28 Oct 2017, 17:37
What is the best way to ensure that l've considered all the possible scenarios? This time l didn't realize {4,4,6,6} is one of the possibilities.

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Re: If four dice are thrown simultaneously,what is the probability that   [#permalink] 28 Oct 2017, 17:37
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