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Number theory Difficulty 1.4 Junior Find the answer Canada

A number is randomly chosen from the list $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , $9$ . The probability that the chosen number
is divisible by $2$ , or by $3$ , or by both $2$ and $3$ , is

Pick one

Solution

From the given list, the numbers that are divisible by $2$ are $2$ , $4$ , $6$ , $8$ .

The numbers that are divisible by $3$
are $3$ , $6$ , $9$ .

The numbers that are divisible by both $2$ and $3$ are the numbers which appear in both of
the previous two lists. The only number in both lists is $6$ .

Thus, the numbers that are divisible by $2$ , or by $3$ , or by both $2$ and $3$ are $2$ , $3$ , $4$ , $6$ , $8$ , $9$ .

Since $2$ , $3$ , $4$ , $6$ , $8$ , $9$
are $6$ of the $9$ numbers listed, then the probability
that the chosen number is divisible by $2$ , or by $3$ , or by both $2$ and $3$ , is $\frac69$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.