Question

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

Answer 1

Let A be event where the first throw comes up with the number 3.
Also, let B be the event where the second throw comes up with the number 3.
∴ Favourable events of A = {(3, 1),(3, 2), (3, 3), (3, 4), (3, 5), (3, 6)} = 6
$\Rightarrow P\left(A\right)=\frac{6}{36}=\frac{1}{6}$
∴ Favourable events of B = {(1, 3), (2, 3), (3, 3), (4, 3),(5, 3), (6, 3) } = 6
$\Rightarrow P\left(B\right)=\frac{6}{36}=\frac{1}{6}$
Also, $P\left(A\cap B\right)=\frac{1}{36}$ [∵ Only one event will be common, i.e. (3, 3)]
Hence, required probability =$P(A\cup B)$
$$\begin{gathered} =P\left(A\right)+P\left(B\right)-P(A\cap B) \\ =\frac{1}{6}+\frac{1}{6}-\frac{1}{36} \\ =\frac{6+6-1}{36} \\ =\frac{11}{36} \end{gathered}$$