# Rolling $4$ dice and multiplying the results. What is the probability that the product is divisible by $5$ or has $5$ as the least significant digit?

Four fair dice are rolled and the four numbers shown are multiplied together. What is the probability that this product

(a) Is divisible by $$5$$?

(b) Has last digit $$5$$?

MY ATTEMPT

a) A number is divisible by $$5$$ if and only if it has at least one factor equal to $$5$$. Let us denote by $$F$$ the event "it occurs at least one number five". Hence the sought probability is given by \begin{align*} \textbf{P}(F) = 1 - \textbf{P}(F^{c}) = 1 - \frac{5^{4}}{6^{4}} = 1 - \left(\frac{5}{6}\right)^{4} \end{align*}

b) Here is the problem. I am not able to describe properly the target results.

Am I on the right track? Can someone please help me to solve it? Any help is appreciated.

• List out all the possibilities for two dice, look for what the pattern is, and then generalize that for four dice. – Acccumulation Jan 11 at 20:48
• The title allows for 150 characters. Please try to use descriptive titles in the future. – Asaf Karagila Jan 11 at 22:49
• Ok, Asaf. Thanks for the tip. – user1337 Jan 11 at 22:53

As others have noted, for part (b) you can first show that:

the product has last digit $$5$$ if and only if all the numbers are odd and there is at least one $$5$$.

From there, I would proceed like this (just slightly simpler than pwerth and José Carlos Santos's approaches):

\begin{align*} P(\text{all odd AND at least one 5}) &= P(\text{all odd}) - P(\text{all odd and no 5}). \end{align*}

So, what is the probability that all dice are odd? And what is the probability all are odd and there is no $$5$$ (i.e., all dice are $$1$$ or $$3$$)?

• +1 because it's the approach I was going to suggest and now I don't need to! (And it must surely be the simplest). – timtfj Jan 11 at 19:14
• If I got it right, the answer is $(1/2)^{4} - (1/3)^{4}$. – user1337 Jan 11 at 19:20
• @user1337 Precisely! – 6005 Jan 11 at 19:20

a) looks good.

b) Hint: A number ends in $$5$$ if it both

• is divisible by $$5$$
• is odd (i.e. not divisible by $$2$$)

Your answer for part $$(a)$$ is correct. For part $$(b)$$, observe that a number $$N$$ has last digit $$5$$ iff it's divisible by $$5$$ and odd. This means that if we denote our rolls by $$a,b,c,d$$ then $$N=abcd$$ where at least one of $$a,b,c,d$$ is a $$5$$ and none of $$a,b,c,d$$ is even. This means we have to avoid $$2,4,6$$ on all rolls, as well as obtain at least one $$5$$.

• Case 1: Exactly one $$5$$. Choose one of $$4$$ positions for the $$5$$ and fill in the remaining three spots with either $$1$$ or $$3$$. This can be done in $$\binom{4}{1}\cdot 2^{3}$$ ways.
• Case 2: Exactly two $$5$$s. Choose two of $$4$$ positions for the $$5$$s and fill in the remaining two spots with either $$1$$ or $$3$$. This can be done in $$\binom{4}{2}\cdot 2^{2}$$ ways.
• Case 3: Exactly three $$5$$s. Choose three of $$4$$ positions for the $$5$$s and fill in the remaining spot with either $$1$$ or $$3$$. This can be done in $$\binom{4}{3}\cdot 2^{1}$$ ways.
• Case 4: All $$5$$s. There is only one way to do this.

The number of desirable outcomes is therefore $$\binom{4}{1}\cdot 2^{3}+\binom{4}{2}\cdot 2^{2}+\binom{4}{3}\cdot 2+ 1$$ and since there are $$6^{4}$$ total outcomes, the desired probability is $$\frac{\binom{4}{1}\cdot 2^{3}+\binom{4}{2}\cdot 2^{2}+\binom{4}{3}\cdot 2+ 1}{6^{4}}$$

Concerning the other question, note that the last digit will be $$5$$ if and only if there is at least one $$5$$ and, besides, all others are odd. So, the answer is the sum of $$4$$ numbers:
• the odds that there's exactly one five and the others are odd numbers: $$\displaystyle4\times\frac16\times\left(\frac13\right)^3$$;
• the odds that there are exactly two fives and the others are odd numbers: $$\displaystyle6\times\left(\frac16\right)^2\times\left(\frac13\right)^2$$;
• the odds that there are exactly three fives and the other is an odd number: $$\displaystyle4\times\left(\frac16\right)^3\times\frac13$$;
• the odds that there are four fives: $$\displaystyle\left(\frac16\right)^4$$.