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?

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