Sure, let's break down each part of the problem and find the probabilities step-by-step:
### (a) The number showing is a 5
A standard die has six faces, numbered from 1 to 6. Only one of these faces shows the number 5.
[tex]\[ P(5) = \frac{\text{Number of outcomes showing a 5}}{\text{Total number of outcomes}} \][/tex]
Since there is only one face with a 5:
[tex]\[ P(5) = \frac{1}{6} \][/tex]
So, the probability that the number showing is a 5 is:
[tex]\[ P(5) = \frac{1}{6} \][/tex]
### (b) The number showing is an even number
The even numbers on a die are 2, 4, and 6. There are three even numbers out of six total possible outcomes.
[tex]\[ P(\text{even}) = \frac{\text{Number of even outcomes}}{\text{Total number of outcomes}} \][/tex]
There are 3 even numbers (2, 4, 6) and 6 total numbers:
[tex]\[ P(\text{even}) = \frac{3}{6} \][/tex]
Simplifying this fraction:
[tex]\[ P(\text{even}) = \frac{1}{2} \][/tex]
So, the probability that the number showing is an even number is:
[tex]\[ P(\text{even}) = \frac{1}{2} \][/tex]
### (c) The number showing is greater than 4
The numbers greater than 4 on a die are 5 and 6. There are two such numbers out of six total possible outcomes.
[tex]\[ P(\text{greater than 4}) = \frac{\text{Number of outcomes greater than 4}}{\text{Total number of outcomes}} \][/tex]
There are 2 numbers greater than 4 (5, 6) and 6 total numbers:
[tex]\[ P(\text{greater than 4}) = \frac{2}{6} \][/tex]
Simplifying this fraction:
[tex]\[ P(\text{greater than 4}) = \frac{1}{3} \][/tex]
So, the probability that the number showing is greater than 4 is:
[tex]\[ P(\text{greater than 4}) = \frac{1}{3} \][/tex]
### Summary:
[tex]\[
\begin{aligned}
(a) & \quad P(5) = \frac{1}{6} \\
(b) & \quad P(\text{even}) = \frac{1}{2} \\
(c) & \quad P(\text{greater than 4}) = \frac{1}{3} \\
\end{aligned}
\][/tex]
