Question

Two fair dice are thrown, find the probability that sum of the points on their uppermost faces is:
(i) a perfect square or divisible by 4.
(ii) greater than 10 or an odd number.

Answer 1

Let S be the sample space.
Thus, we have:
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Or,
n(S) = 36

(i) Let A be the event that the sum of the points on the uppermost faces is a perfect square and B be the event that sum of the points on the uppermost faces is divisible by 4.
Thus, we have:
A = {(1,3), (2,2), (3,6), (4,5), (5,4), (6, 3)}
B = {(1,3), (2,2), (2,6), (3,1), (3,5), (4,4), (5,3), (6,2), (6,6)}
A $\cap$B = {(1, 3), (2, 2)}
n(A) = 6, n(B) = 9 and n(A $\cap$B) = 2
And,
$$\begin{gathered} \mathrm{P}\left(\mathrm{A}\right)=\frac{n\left(\mathrm{A}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6} \\ \mathrm{P}\left(\mathrm{B}\right)=\frac{n\left(\mathrm{B}\right)}{n\left(\mathrm{S}\right)}=\frac{9}{36}=\frac{1}{4} \\ \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{n(\mathrm{A}\cap \mathrm{B})}{n\left(\mathrm{S}\right)}=\frac{2}{36}=\frac{1}{18} \end{gathered}$$
Now, on applying the addition theorem of probability, we get:

P(A $\cup$ B) = P(A) + P(B) $-$ P(A $\cap$ B)
$$\begin{gathered} =\frac{1}{6}+\frac{1}{4}-\frac{1}{18} \\ =\frac{6+9-2}{36} \\ =\frac{13}{36} \end{gathered}$$

(ii) Let C be the event that the sum of the points on the uppermost faces is greater than 10 and D be the event that the sum of the points on the uppermost faces is an odd number.
Thus, we have:
C = {(5,6), (6,5), (6,6)}
D = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}
C $\cap$ D = {(5, 6), (6, 5)}
n(C) = 3, n(D) = 18 and n(C $\cap$ D) = 2
And,
$$\begin{gathered} \mathrm{P}\left(\mathrm{C}\right)=\frac{n\left(\mathrm{C}\right)}{n\left(\mathrm{S}\right)}=\frac{3}{36}=\frac{1}{12} \\ \mathrm{P}\left(\mathrm{D}\right)=\frac{n\left(\mathrm{D}\right)}{n\left(\mathrm{S}\right)}=\frac{18}{36}=\frac{1}{2} \\ \mathrm{P}(\mathrm{C}\cap \mathrm{D})=\frac{n(\mathrm{C}\cap \mathrm{D})}{n\left(\mathrm{S}\right)}=\frac{2}{36}=\frac{1}{18} \end{gathered}$$
Now, on applying the addition theorem of probability, we get:

P(C $\cup$ D) = P(C) + P(D) $-$ P(C$\cap$ D)
$$\begin{gathered} =\frac{1}{12}+\frac{1}{2}-\frac{1}{18} \\ =\frac{3+18-2}{36}=\frac{19}{36} \end{gathered}$$