You roll a die 8 times. What is the probability of getting an odd number at least twice and an even number at least three times? My initial idea was to do something such as $3^5 * 6^3$ for the possible cases. However, I believe this is not appropriate for this question, since simply multiplying is usually used in cases in which order matters (e.g. combination of a locker), and here order does not matter.
I'd imagine that maybe we should use something like Bernoulli trials or Combinations, but I'm not sure.
Note: I'm not saying this the the answer to your question. I'm showing you the binomial distribution and a process for adding desired
terms. You need to decide which terms are
needed to answer your question.
The probability of an Odd result on any one roll is $p=1/2.$ The number $X$ of odd results in $n = 8$ rolls is
distributed $X \sim \mathsf{Binom}(n = 8, p = 1/2).$
Specifically, the probability that $X = k$ is
$$P(X = k) = {8\choose k}(1/2)^8,$$
for $k = 0,1,\dots,8.$
Suppose you want the probability $P(2 \le X \le 5).$ That's
$$P(2 \le X \le 5) = P(X \le 5) - P(X \le 1)\\ 
= \sum_{k=2}^5 {8\choose k}(1/2)^8 = 0.8203,$$
to four places. To get this probability, you have four terms to add.
In R statistical software this can be computed
as follows, where dbinom is a binomial PDF and pbinom is a binomial CDF.
k = 2:5; dbinom(k, 8, 1/2)
[1] 0.1093750 0.2187500 0.2734375 0.2187500
sum(dbinom(k, 8, 1/2))  # add 4 PDF terms
[1] 0.8203125

pbinom(1:5, 8, 1/2)
[1] 0.03515625 0.14453125 0.36328125 0.63671875 0.85546875
diff(pbinom(c(1,5), 8, 1/2))  # subtract 2 CDF values
[1] 0.8203125

In the figure below, we are adding the heights of the four
maroon bars.

R code for figure above:
plot(0:8, dbinom(0:8, 8, 1/2), type=&quot;h&quot;, lwd=2,
     col=&quot;blue&quot;, ylab=&quot;PDF&quot;, xlab=&quot;k&quot;, 
     main=&quot;PDF of BINOM(8, .5)&quot;)
 abline(h = 0, col=&quot;green2&quot;)
 lines(2:5, dbinom(2:5, 8, 1/2), type=&quot;h&quot;, 
       lwd=2, col=&quot;maroon&quot;)
The probability of getting  an odd number at least 2 times = the probability of getting an odd number exactly 2 times + probability of getting odd number 3 times
$n$=8.
Hint: Probability of getting an odd number exactly twice:
$8*(\frac{1}{2})^2$*$\frac{1}{2}$.