Question

In a single throw of three dice, determine the probability of getting
i) a total of $5$
ii) a total of atmost $5$
iii) a total of at least $5$

Answer 1

Simple space $=6\times 6\times 6=216$

Total outcomes $=216$

$\left(i\right)$Total of $5$

$a+b+c=5$

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

$\therefore$ probability $=\frac{6}{216}=\frac{1}{36}$

$\left(ii\right)$ Atotal of atmost $5=(1,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1),(1,1,2),(2,1,1),(1,2,1)$

$\therefore$ prob $=\frac{10}{216}=\frac{5}{108}$

$\left(iii\right)$Probability of getting atleast $5=P\left(5\right)-P(<5)$

$=1-\frac{4}{216}=\frac{212}{216}=\frac{53}{54}$