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 $5$?


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.

  • 1
    $\begingroup$ List out all the possibilities for two dice, look for what the pattern is, and then generalize that for four dice. $\endgroup$ – Acccumulation Jan 11 at 20:48
  • $\begingroup$ The title allows for 150 characters. Please try to use descriptive titles in the future. $\endgroup$ – Asaf Karagila Jan 11 at 22:49
  • $\begingroup$ Ok, Asaf. Thanks for the tip. $\endgroup$ – 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
    $\begingroup$ +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). $\endgroup$ – timtfj Jan 11 at 19:14
  • $\begingroup$ If I got it right, the answer is $(1/2)^{4} - (1/3)^{4}$. $\endgroup$ – user1337 Jan 11 at 19:20
  • $\begingroup$ @user1337 Precisely! $\endgroup$ – 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 $$\frac{\binom{4}{1}\cdot 2^{3}+\binom{4}{2}\cdot 2^{2}+\binom{4}{3}\cdot 2+ 1}{6^{4}}$$


Your answer to the first question is correct.

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: $\displaystyle\left(\frac16\right)^4$.

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