Answer
(a) The probability of the event that the result is a repeated number is $\frac{1}{3}$.
(b) The probability of the event that the second number is 1 or 3 is $\frac{2}{3}$.
(c) The probability of the event that the first number is even and the second number is odd is $\frac{2}{9}$.
Work Step by Step
The sample space is
$S=${$1,1;1,2;1,3;2,1;2,2;2,3;3,1;3,2;3,3$}, therefore the total number of outcomes is 9.
The probabitity of an event can be calculated as:
$P(A)=\frac{\text{favorable outcomes out of the sample set}}{\text{total number of possible outcomes}}$
This means, that the probability of the event that the result is a repeated number is $\frac{1}{3}$, as there are three favorable outcomes $(1,1;2,2;3,3)$ out of the sample:.
$P(A)=\frac{3}{9}=\frac{1}{3}$
The probability of the event that the second number is 1 or 3 is $\frac{2}{3}$, as there are six favorable outcomes $(1,1;1,3;2,1;2,3;3,1;3,3)$ out of the sample space:
$P(B)=\frac{6}{9}=\frac{2}{3}$
(
The probability of the event that the first number is even and the second number is odd is $\frac{2}{9}$, as there are two favorable outcomes $2,1;2,3)$ out of the sample space:
$P(B)=\frac{2}{9}$
