Question

A die is thrown twice. What is the probability that at least one of the two throws come up with the number 3?

Answer 1

A die is thrown twice

$\Rightarrow n\left(S\right)=36$

Let A be the event of getting 3 first throw

$\therefore P\left(A\right)=\frac{6}{36}=\frac{1}{6}$

[$\because$ A={(3,1),(3,2),(3,3),(3,4),(3,5)}]

B be the event of getting 3 in 2nd throw

$P\left(B\right)=\frac{6}{36}=\frac{1}{6}$

Also, $P(A\cap B)=\frac{1}{36}$

$[\because A\cap B=\{3,3\}]$

$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

= $\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$