I came across this problem in a book that I bought recently.
Three dice are dropped. Find the probability that when we roll the dice:
a) In all of the dice we score $5$. (Not as a sum.)
My answer: $\frac3 {18} = \frac16$.
b) In at least one die we score $6$.
My answer: $\frac3{18}=\frac16$.
c) The sum of the scores in the dice is smaller or equal than $6$.
My answer: since we have the combinations $(1 1 1), (1 1 2), (1 1 3),(1 1 4),(2 1 2), (2 1 3), (4 1 1)$, we arrive at $\frac7{18}$.
d) The sum of the scores in the dice is greater than the product of the scores.
I don’t know how to do this one.
e) We don’t score neither $5$ nor $6$ in any of the dice.
This has to be $\frac3{18}+\frac3{18} =\frac6{18}=\frac13$.
f) In two dice, we have the same score.
I don’t know how to do this one either.
To start off, you will need to understand the idea of sample space of this experiment. There are $6$ different ways in which each die can turn out. Hence, the total number of possibilities, i.e. the sample space is $6^3 = 216$ possibilities.
a) If we score $5$ in all of the three dice, there is only $1$ way of the $216$ ways. Hence, the final answer is $\frac 1 {216}$
b) Consider if I get the first die to be $6$. There are $36$ ways, i.e. the possibilities are $(6, *, *)$ where $*$ represents any number from $1$ to $6$. There are also $36$ ways respectively for $(*,6,*)$ and $(*, *, 6)$. However, you will need to remove the cases of two $6$ appearing i.e. $(6, 6,*), (6,*,6) \&(*,6,6)$ with $6$ cases each. Finally, add back the case $(6,6,6)$.The final answer will be
$$\text{Req. Probability} = 3 \times\frac{36}{216} - 3 \times \frac{6}{216} + \frac{1}{216}$$
c) I would recommend you to plot a matrix for this one. I will only plot the pairs of dice that add up to $5$ (keep in mind there's one more die in play)
\begin{bmatrix} (1,1) & (1,2) & (1,3) & (1,4) \\ (2,1) & (2,2) & (2,3) \\ (3,1) & (3,2)\\ (4,1) \end{bmatrix}
For each tuple, there are the following number of different cases where the third die can show up: \begin{bmatrix} 4 & 3 & 2 & 1\\ 3 & 2 & 1\\ 2 & 1\\ 1 \end{bmatrix} Hence, the final answer is
$$\text{Req. Probability} = \frac{4+3+2+1+3+2+1+2+1+1}{216}.$$