Three fair 6 sided dice are rolled. Find the probability that the total on all three dice is 5 or less.?
ans10/216 pleease explain your method :-)
Hello, aimsywamesy!
Three fair 6-sided dice are rolled.
Find the probability that the total on all three dice is 5 or less.
Ans: 10/216
There are: $\displaystyle 6^3 = 216$ possible outcomes.
Outcomes with sums $\displaystyle \le 5$
. . $\displaystyle \begin{array}{c} (1,1,1) \\ (1,1,2)\;(1,2,1)\;(2,1,1) \\ (1,1,3)\;(1,3,1)\;(3,1,1) \\ (1,2,2)\;(2,1,2)\;(2,2,1) \end{array}$ . . 10 outcomes
Therefore: .$\displaystyle P(\text{sum}\le 5) \;=\;\frac{10}{216} \;=\;\frac{5}{108} $
There are 216 possible triples.
Make a list of favorable outcomes.
We can get 1,1,1 in one way.
We can get 1,1,2 in three ways: 112, 121, 211.
Same for 113. So far that is seven favorable outcomes.
According to the give answer you should find ten in all.
Or go to this webpage add up the coefficients of $\displaystyle x^3,~x^4,~\&~x^5$.