If 3 six faced fair dice are thrown together, then the probability that the sum of the numbers appearing on the dice is k where $ 9\leqslant k \leqslant 14$ is :
My approach : Possible sum values are from 3 to 18 and our cases of interst are $9,10,11,12,13,14 \therefore$ probability should be $\frac{6}{16}$ but the answer given is $\frac{21k-k^{2}-83}{216}$ How can arrive at the given answer ?
Best Answer
Consider only Sum of all outcomes to be 9. Then see the cases where when 3 fair dices are thrown, the sum comes out to be 9. There will be 25 such cases when checked.
Means probability will be 25/216.
With this knowledge, put k=9 in the given equations. The equation which will give the same output is the answer. Thus, equation with 21k-k^2-83 as numerator satisfies our requirement.