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Three distinct numbers are selected at random from the set $\{1,2,3,4,5,6\}$. What is the probability that their product is divisible by $3$?

I think that since because $3$ and $6$ are the only numbers that divide into three with an integer remainder, the number would have to have $3$ or $6$ in the unit's digit. Since this is the case the probability would be $2/6$ or $1/3$. Have I done anything wrong?

amWhy
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Led Nowmann
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  • You are correct that to be a product of 3, one of the terms must be 3 or 6. What I don't understand is how you got that the probability of that is 2/6. Work on that. – fleablood Feb 19 '18 at 01:44
  • There are ${6\choose 3} $ ways to pick 3 elements. ${4\choose 3} $ of those have neither 3 nor 6. So ${6\choose 3}-{4\choose 6} $ have a 3 or a 6 or both. So probability is $\frac {{6\choose 3}-{4-3}}{6\choose 3}$ – fleablood Feb 19 '18 at 01:49

1 Answers1

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We need to find the probability that at least one of the numbers is divisible by three. This is the same as

$$\begin{align*} 1-P(\text{none of the numbers are divisible by three}) &=1-\left(\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\right)\\\\ &=1-0.2\\\\ &=0.8 \end{align*}$$

In order to get three numbers that are not $3$ or $6$, we must first select $1,2,4,$ or $5$ out of the six numbers giving a probability of $\frac{4}{6}$.

Take that number away. We must now select one of the remaining three numbers of the five with probability $\frac{3}{5}$.

Finally, we must now select one of the two remaining numbers of the four with probability $\frac{2}{4}$

Remy
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