From a table of random numbers, one number is selected at…

From a table of random numbers, one number is selected at random. If the selected number is divisible by $5$, this is event $A$; if the number ends with zero, this is event $B$. What does the event $A\setminus B$ mean?

Step 1: The difference of events

For events $A$ and $B$ the difference is

\[ A\setminus B = \{\omega : \omega\in A \text{ and } \omega\notin B\} = A\overline{B}. \]

In words: "$A$ occurs, but $B$ does not occur".

Step 2: Describe $A$ and $B$ by the last digit

Numbers in a table of random numbers are non-negative integers. Any such number can be written as $n = 10q + c$, where $c\in\{0,1,\dots,9\}$ is its last digit. Since $10q$ is always divisible by $5$, the number $n$ is divisible by $5$ exactly when its last digit $c$ is divisible by $5$, i.e. $c = 0$ or $c = 5$.

• $A$ = "divisible by $5$" = "the last digit is $0$ or $5$": $\ 0, 5, 10, 15, 20, 25, \dots$
• $B$ = "ends with zero" = "the last digit is $0$": $\ 0, 10, 20, 30, \dots$

Step 3: $B$ is contained in $A$

If a number ends with zero, it has the form $10q = 5\cdot(2q)$, so it is divisible by $5$. Hence every outcome of $B$ is also an outcome of $A$:

\[ B\subseteq A. \]

But $A$ is larger than $B$: it also contains the numbers ending with $5$.

Step 4: Find $A\setminus B$

A number belongs to $A\setminus B$ when its last digit is $0$ or $5$ (it is in $A$) and its last digit is not $0$ (it is not in $B$). Only the digit $5$ remains:

\[ \boxed{A\setminus B = \text{"the selected number ends with the digit } 5\text{"}.} \]

Equivalent descriptions of the same event:

• the number is divisible by $5$ but not by $10$;
• the number is an odd multiple of $5$;
• the number has the form $10q + 5$, $q = 0, 1, 2, \dots$

Examples: $5,\ 15,\ 25,\ 35,\ 45,\ \dots,\ 105,\ 115,\ \dots$

Step 5: Check

Since $B\subseteq A$, the event $A$ splits into two non-overlapping parts:

\[ A = B + (A\setminus B): \quad \{\text{ends with } 0 \text{ or } 5\} = \{\text{ends with } 0\} + \{\text{ends with } 5\}. \ \checkmark \]

Step 6: Probabilities (bonus)

In a table of random numbers every digit $0,1,\dots,9$ appears in the last place with the same probability $\tfrac{1}{10}$. Therefore

\[ P(A) = \frac{2}{10} = 0.2, \qquad P(B) = \frac{1}{10} = 0.1, \qquad P(A\setminus B) = \frac{1}{10} = 0.1. \]

Because $B\subseteq A$, we also have $P(A\setminus B) = P(A) - P(B) = 0.2 - 0.1 = 0.1$, in agreement.