5
$\begingroup$

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.

$\endgroup$
  • 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
8
$\begingroup$

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 $3$)?

$\endgroup$
  • 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
5
$\begingroup$

a) looks good.

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

  • is divisible by $5$
  • is odd (i.e. not divisible by $2$)
$\endgroup$
4
$\begingroup$

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 $$\frac{\binom{4}{1}\cdot 2^{3}+\binom{4}{2}\cdot 2^{2}+\binom{4}{3}\cdot 2+ 1}{6^{4}}$$

$\endgroup$
3
$\begingroup$

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.