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 = \[P(A \cup B)\]

\[= P(A) + P(B) - P(A \cap B)\]
\[ = \frac{1}{6} + \frac{1}{6} - \frac{1}{36}\]
\[ = \frac{6 + 6 - 1}{36}\]
\[ = \frac{11}{36}\]