# Is this interpretation correct for this probability question?

Seven chits are numbered $$1$$ to $$7$$. Three are drawn one by one with replacements. The probability that the least number on any selected chit is $$5$$, is

My teacher says that the favourable outcomes are $$5, 6, 7$$, so the probability is $$\left( \frac 3 7 \right)^3$$.

But I think that this also includes cases like $$(6, 6, 7)$$. So I'm getting the probability of $$\frac{3^3-2^3}{7^3}$$, after excluding the cases of only $$6$$'s and $$7$$'s.

Which of these is the correct interpretation of the question? Thanks.

• Why are you excluding cases which have only $6$ or $7$? Jul 15, 2021 at 4:28
• @Math Lover If the least number is $5$, then a $5$ must be present. Jul 15, 2021 at 5:45
• @DanielMathias that is not how I would interpret. The least number on any selected chit is $5$ means it can be either $5, 6$ or $7$. Jul 15, 2021 at 6:01
• I agree that there is ambiguity. The solution depends upon whether "The probability that the least number on any selected chit is 5" means the least value of the three draws is exactly $5$ or if we are looking for the probability that each selection has a value that is at least $5$. Jul 15, 2021 at 6:07
• @Math Lover $x=5$ does not mean $x\ge 5$ There is absolutely no ambiguity here. Jul 15, 2021 at 9:02

The event was that the least number that shows on any of the three draws is $$5$$.
The event was not that $$5$$ is the smallest number showing on the three draws.
There is no requirement that $$5$$ shows on any draw; only that nothing less than that ever shows.
• If there is no $5$, then $5$ cannot be the least number shown. Jul 15, 2021 at 5:47