Animated Solution for Mathematics - Probability: An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

Imagine you are standing in front of a bag containing fifty slips of paper, each numbered from 1 to 50. You are asked to pick one at random, and you want to know the probability that the number you pick is a multiple of 4, 6, or 7.

We define our sample space $S=\{1,2,\dots ,50\}$, where the total number of outcomes is $n(S)=50$. To solve this, we define three sets: $A$ for multiples of 4, $B$ for multiples of 6, and $C$ for multiples of 7.

To count the multiples of a number $k$ up to 50, we use the floor function $\lfloor \frac{50}{k}\rfloor$.

For set $A$, we calculate:

For set $B$, we find:

For set $C$, we find:

Some numbers are multiples of both 4 and 6, such as 12 or 24. These numbers have been counted in both set $A$ and set $B$. To correct this, we must find the intersection of these sets using the Least Common Multiple ($\text{LCM}$).

For $A$ and $B$, $\text{LCM}(4,6)=12$, so:

For $B$ and $C$, $\text{LCM}(6,7)=42$, so:

For $A$ and $C$, $\text{LCM}(4,7)=28$, so:

For the triple intersection $n(A\cap B\cap C)$, we look for multiples of $\text{LCM}(4,6,7)=84$. Since 84 is greater than 50, there are zero such numbers:

We now apply the Principle of Inclusion-Exclusion to find the union $n(A\cup B\cup C)$. The formula is:

Substituting our values, we get:

We have identified 21 favorable outcomes.

The probability $P$ is the ratio of favorable outcomes to total outcomes.

This elegant result shows how structured thinking can turn a complex counting problem into a simple, satisfying solution. The final probability is 0.42.