What is the probability of getting a sum of 12 when rolling

Sum of 12 can be achieved in following ways
6,5,1---Total cases = 3! = 6
6,4,2---Total cases = 3!= 6
6,3,3---Total cases = 3!/2! = 3
5,5,2---Total cases = 3!/2! = 3
5,4,3---Total cases = 3! = 6
4,4,4---Total cases = 3!/3! = 1

Total cases = 25

Probability = 25 * (1/6 * 1/6 * 1/6) = 25/216
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I broke my head 15 minutes on it, then the soln is so devilishly simple!

Answer is E > 25/100

Basically, if the numbers on two of the dice are decided, then there is only one value of the third die that is possible. So you have to find only those combinations of two dice that yield a sum of 6 or above. e.g. If the value on the 1st die is 1, then you have two possible outcomes - 5 or 6 on the 2nd die. Similarly, for a value of 2 on the first die, you have 3 possible outcomes (4, 5, 6) and so on... However, if you get a value of 6 on the first die, then you can have only 5 possible outcomes.

The total number of outcomes comes out to be (2 + 3 + 4 + 5 + 6 + 5) = 25.

Probability is 25/216!
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let the dices be a,b,c

a+b+c = 12

but 0 < a,b,c < 6

so we can re-write the equation as

(6-a)+(6-b)+(6-c) = 12
=> a+b+c = 6
which has total of C(8,2) whole number solutions = 28
but out of these 28, three cases must be where 2 of a,b, or c is 0 and the other is 6. so we need to remove those cases which would be 3 cases ( C(3,2) )
so total 25 such cases

answer = 25/216.
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I think it is very easy...........
Start checking from the smaller or bigger numbers on the dice. We will check from bigger numbers working downwards: start with 6, it has the following options: (6,5,1), (6,4,2), (6,3,3). Now pass on to 5: (5,5,2), (5,4,3). Now 4: (4,4,4). And that’s it, these are all number combinations that are possible, if you go on to 3, you will notice that you need to use 4, 5 or 6, that you have already considered (the same goes for 2 and 1). Now analyze every option: 6,5,1 has 6 options (6,5,1), (6,1,5), (5,1,6), (5,6,1), (1,6,5), (1,5,6). So do (6,4,2) and (5,4,3). Options (6,3,3) and (5,5,2) have 3 options each: (5,5,2), (5,2,5) and (2,5,5). The same goes for (6,3,3). The last option (4,4,4) has only one option. The total is 3*6+2*3+1=18+6+1 = 25 out of 216 (6x6x6) options.

5,5,2= (552),(255),(525)
Like= 336

An easy way to visualise it is by drawing a Binomial tree, we've all done those in school and if each branch has a 1/2 probability, you'll find that there's only 3 ways to get there.

(1/2)^3 x 3 = 3/8


Hope it helps!

G

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Excellent!!! Thanks Karishma

Thanks Karishma,

Could you please explain how 9 is calculated that is subtracted from 9C2 in below scenario?

Similarly, how will you adjust for the sum of 10? There will be cases where the split is like this: (7, 0, 0) or (6, 1, 0). These do not work since the maximum a die can show is 6. So you need to remove 9 cases (3 arrangements of 7, 0, 0 and 6 arrangements of 6, 1, 0)

I understand using this method and have used. But here I didn't understand the rewriting the equation part. Please explain that and what follows that. Bunuel, if possible, would be great if you could help me out here.

Given

• Three dices are rolled simultaneously.

To Find

• The probability of getting a sum of 12.

Approach and Working Out

• 12 can come in the following ways,

o 6, 5, 1 – 3! ways
o 6, 4, 2 – 3! Ways
o 6, 3, 3 – 3!/2! ways
o 5, 5, 2 – 3!/2! ways
o 5, 4, 3 – 3! ways
o 4, 4, 4 – 1 way
o Total ways = 6 + 6 + 3 + 3 + 6 + 1 = 25 ways

• Total cases \(6^3\) = 216 ways

• Answer = \(\frac{25}{216}\)

Correct Answer: Option D
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The total outcome in a roll of a die: 6

Overall outcomes rolling three dice simultaneously = \(6^3\) = 216

The desired outcome of sum = 12

If the first die has 6, the rest two of the dice needs 6: [1,5],[2,4],[3,3].

=> For instance if numbers are 6, 1, 5, this can be pair up in 3! ways = 6 except 6,3,3 [\(\frac{3!}{2!}\) = 3 ways].

So, total ways: 6 * 2 + 3 = 15

If the first die has 5, the rest two of the dice needs 7: [1,6],[2,5],[3,4]: [1,6,5] is already considered, 5 with [2,5] will be arranged in \(\frac{3!}{2!}\) = 3 ways and 5 with [3,4] in 6 ways = 3 + 6 = 9 ways

If the first die has 4, the rest two of the dice needs 8: [2,6],[3,5],[4,4]: [2,6,4] is already considered, 4 with [3,5] is already considered and 4 with [4,4] in 1 way = 1 way

If the first die has 3, the rest two of the dice needs 9: [3,6],[4,5]: Both cases already considered.

Overall results: 15 + 9 + 1 = 25

Probability: \(\frac{25}{216}\)

Answer E

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