Question

A die is thrown twice. Find the probability that at least one of the two throws comes up with the number $5$.

Answer 1

We have, $S=\left\{\right(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),......(6,1),(6,2),(6,3),(6,4),(6,5),(6,6\left)\right\}$
Thus, $n\left(S\right)=36$
Let $A$ be the event of getting $5$ in the first throw.
$\Rightarrow A=\left\{\right(5,1),(5,2),(5,3),(5,4),(5,5),(5,6\left)\right\}$
$\Rightarrow n\left(A\right)=6$
$\Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{26}$
Let $B$ be the event of getting $5$ in the second throw
$\Rightarrow B=\left\{\right(1,5),(2,5),(3,5),(4,5),(5,5),(5,6\left)\right\}$
$\Rightarrow n\left(B\right)=6$
$\Rightarrow P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{6}{36}$
Thus, $A\cap B=\left\{\right(5,5\left)\right\}$
and $n(A\cap B)=1$
Therefore, $P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{1}{36}$
The required probability $=P(A\cup B)=$
$=P\left(A\right)+P\left(B\right)-P(P(A\cap B)=\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}$