Answer
(a) The probability of the event that both slips are marked with even numbers is $\frac{1}{6}$
(b) The probability of the event that both of the slips are marked with odd numbers is $\frac{1}{6}$.
(c) The probability of the event that both of the slips are marked with the same number is $0$.
(d) The probability of the event that one slip is marked with an odd number, and the other one with an even number is $\frac{2}{3}$.
Work Step by Step
The sample set is
$S=${$4,3;4,2;4,1;3,4;3,2;3,1;2,4;2,3;2,1;1,4;1,3;1,2$}, therefore the total number of outcomes is 12.
The probalitity 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 both slips are marked with even numbers is $\frac{1}{6}$, as there are two favorable outcomes $(2,4;4,2)$ out of the sample set:
$P(A)=\frac{2}{12}=\frac{1}{6}$
The probability of the event that both of the slips are marked with odd numbers is $\frac{1}{6}$, as there are two favorable outcomes $(1,3;3,1)$ out of the sample set:
$P(B)=\frac{2}{12}=\frac{1}{6}$
The probability of the event that both of the slips are marked with the same number is 0, as there isn't any favorable outcome out of the sample set:
$P(B)=\frac{0}{12}=0$
The probability of the event that one slip is marked with an odd number, and the other one with an even number is $\frac{2}{3}$, as there are eight favorable outcomes $(1,2;1,4;2,1;2,3;3,2;3,4;4,1;4,3)$ out of the sample set:
$P(B)=\frac{8}{12}=\frac{2}{3}$
