A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16?
I have a problem with these type if questions, especially if the sum is sth like 12,etc. (the number of combinations are large)
What is the way to solve this problem?
could somebody solve this using the expected value concept please?
The sum of 16 is obtained if the dices show: (6, 5, 5) or (6, 6, 4).
Each can occur in 3!/2!=3 ways, for example we can get (6, 5, 5) in 3 ways: (6, 5, 5), (5, 6, 5), (5, 5, 6), permutation of three items (6, 6, 5) out of which 2 5's are identical.
So, # of favorable outcomes is 2*3=6;
Total # of outcomes is 6*6*6=216;
As for your question: generally if on the real test you are faced with some hard calculations, something like 1.79*4.89, then it might indicate two cases: either there is a shortcut solution for this problem, which won't involve such tedious math or answer choices are handy to deal with it.
Now, 12 can be obtained by combination of the following 6 cases: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25 --> P=25/216. Since, it involves counting of 6 different cases and there is no real shortcut, then I doubt that you'll be asked about the probability of getting the sum of 12 (or the answers choices will be useful to find this value).
Having said that you should take into account the fact that the probability distribution in this question (and in many similar questions) is symmetrical and sometimes that could give a good shortcut. For example:
The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum);
The probability of getting the sum of 4 = the probability of getting the sum of 17;
The probability of getting the sum of 5 = the probability of getting the sum of 16;
The probability of getting the sum of 10 = the probability of getting the sum of 11;
So, for our case, the probability of getting the sum of 16 is the same as the probability of getting the sum of 5 and you can find the combinations for the sum 5 instead: (2, 2, 1) or (3, 1, 1).
To summarize: combinations/probability questions can be solve in many ways, most of them are fairly straightforward, some have tricky shortcut and some can be solved wit the help of answer choices, so no hard math will be required.
Check similar question which involves expected value approach to practice: mary-and-joe-126407.html
Hope it helps.
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