Question

If two dice are rolled simultaneously, find the probability of the following events.
(1) The sum of the digits on the upper faces is at least 10.
(2) The sum of the digits on the upper faces is 33.
(3) The digit on the first die is greater than the digit on second die.

Answer 1

$\mathrm{Sample}\mathrm{Space}\left(\mathrm{S}\right)=\left\{\begin{array}{cccccc}\left(1,1\right)& \left(1,2\right)& \left(1,3\right)& \left(1,4\right)& \left(1,5\right)& \left(1,6\right)\\ \left(2,1\right)& \left(2,2\right)& \left(2,3\right)& \left(2,4\right)& \left(2,5\right)& \left(2,6\right)\\ \left(3,1\right)& \left(3,2\right)& \left(3,3\right)& \left(3,4\right)& \left(3,5\right)& \left(3,6\right)\\ \left(4,1\right)& \left(4,2\right)& \left(4,3\right)& \left(4,4\right)& \left(4,5\right)& \left(4,6\right)\\ \left(5,1\right)& \left(5,2\right)& \left(5,3\right)& \left(5,4\right)& \left(5,5\right)& \left(5,6\right)\\ \left(6,1\right)& \left(6,2\right)& \left(6,3\right)& \left(6,4\right)& \left(6,5\right)& \left(6,6\right)\end{array}\right\}$
∴ n(S) = 36

(1) Event A: sum of the digits on the upper faces is at least 10
A = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
∴ n(A) = 6
$$\begin{gathered} \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{6}{36} \\ =\frac{1}{6} \end{gathered}$$
Hence, the probability that the sum of the digits on the upper faces is at least 10 is $\frac{1}{6}$.
(2) Event B: sum of the digits on the upper faces is 33
B = {}
∴ n(B) = 0
$$\begin{gathered} \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{n}\left(\mathrm{B}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{0}{36} \end{gathered}$$
= 0
​Hence, the probability that the sum of the digits on the upper faces is 33 is 0.

(3) Event C: digit on the first die is greater than the digit on second die
C = {(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5)}
∴ n(C) = 15
$$\begin{gathered} \therefore \mathrm{P}\left(\mathrm{C}\right)=\frac{\mathrm{n}\left(\mathrm{C}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{15}{36} \\ =\frac{5}{12} \end{gathered}$$
​Hence, the probability that digit on the first die is greater than the digit on second die is $\frac{5}{12}$.