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

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.

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

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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