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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.

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 $3$)?

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: $displaystyle4timesfrac16timesleft(frac13right)^3$;


• the odds that there are exactly two fives and the others are odd numbers: $displaystyle6timesleft(frac16right)^2timesleft(frac13right)^2$;


• the odds that there are exactly three fives and the other is an odd number: $displaystyle4timesleft(frac16right)^3timesfrac13$;


• the odds that there are four fives: $displaystyleleft(frac16right)^4$.