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

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:

GMAT assassins aren't born, they're made,
Rich

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.

T_T

I have the same question.

Additionally, why don't we have to consider the probability of selecting each number as well? @shrouded solution is exactly what I did but I also accounted for the probability of selecting each number

e.g. 6 6 5 5 = (1/6)^2 + (1/6)^2

Can a math expert help here?

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

20 = 2 + 6 + 6 + 6 (4 ways) = 3 + 5 + 6 + 6 (12 ways) = 4 + 4 + 6 + 6 (6 ways) = 4 + 5 + 5 + 6 (12 ways) = 5 + 5 + 5 + 5 (1 way) = 35 ways
Total ways = 6^4 = 1296

The probability that the sum of the number is exactly 20 = 35/1296

IMO A

Can you explain how you calculated the 4 ways, 12 ways, 6 ways, etc.?

Given that four fair dice are thrown simultaneously and We need to find what is the probability that the sum of the number is exactly 20?

As we are rolling four dice => Number of cases = \(6^4\) = 1296

Now we need to find the cases when the sum of 4 outcomes = 20

If we get 6 in three dice then the fourth dice should have a 2 to get the sum as 20
=> None of the dice can have 1 as the outcome.

Let's write all possible cases in which we can get sum of 4 numbers (Between 1 to 6) as 20
=> 2 + 6 + 6 + 6 = 20
=> 3 + 5 + 6 + 6 = 20
=> 4 + 4 + 6 + 6 = 20
=> 4 + 5 + 5 + 6 = 20
=> 5 + 5 + 5 + 5 = 20

{2,6,6,6} we can get this combination in \(\frac{4!}{3!}\) ways (4! because we have 4 numbers and 3! as 6 is getting repeated 3 times) = 4 ways

{3,5,6,6} we can get this combination in \(\frac{4!}{2!}\) ways (4! because we have 4 numbers and 2! as 6 is getting repeated 3 times) = 12 ways

{4,4,6,6} we can get this combination in \(\frac{4!}{2!*2!}\) ways (4! because we have 4 numbers and 2!*2! as 4 and 6 are getting repeated 2 times) = 6 ways

{4,5,5,6} we can get this combination in \(\frac{4!}{2!}\) ways (4! because we have 4 numbers and 2! as 5 is getting repeated 2 times) = 12 ways

{5,5,5,5} we can get this combination in \(\frac{4!}{4!}\) ways (4! because we have 4 numbers and 4! as 5 is getting repeated 5 times) = 1 ways

=> Total number of ways = 4 + 12 + 6 + 12 + 1 = 35

=> Probability that the sum of the number is exactly 20 = \(\frac{35}{1296}\)

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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