Question

In a single throw of three dice, determine the probability of getting
(i) a total of 5.
(ii) total of atmost 5.
(iii) a total of atleast 5.

Answer 1

Total number of outcomes of throwing three dice simultaneously. $n\left(S\right)=6\times 6\times 6=6^3=216$

(i) A total of 5 can be obtained in one of the following ways:

(1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2),(2,2,1)

$\therefore$ Favourable outcomes, $n(f_1=6$

Hence, required probability $=\frac{n\left(f_1\right)}{n\left(S\right)}=\frac{6}{216}=\frac{1}{36}$

(ii) A total of atmost 5 can be obtained in anyone of the following ways:

(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)

$\therefore$ Favourable outcomes, $n\left(f_2\right)=10$

Hence, required prabability $=\frac{n\left(f_2\right)}{n\left(S\right)}=\frac{10}{216}=\frac{5}{108}$

(iii) Let A be the event 'getting atleast 5'.

Then, $P\left(A\right)=1-P\left(\overline{A}\right)=1-P$ (getting a total of atmost 4)

A total of atmost 4 can be obtained in anyone of the following ways:

(1,1,1), (1,1,2), (1,2,1), (2,1,1)

$\therefore P\left(\overline{A}\right)=\frac{4}{216}$

$\therefore P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{4}{216}=\frac{212}{216}=\frac{53}{54}$