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?

Question

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.

Answer 1

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:

$8*(\frac{1}{2})^2$*$\frac{1}{2}$.

Answer 2

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="h", lwd=2,
     col="blue", ylab="PDF", xlab="k", 
     main="PDF of BINOM(8, .5)")
 abline(h = 0, col="green2")
 lines(2:5, dbinom(2:5, 8, 1/2), type="h", 
       lwd=2, col="maroon")

Answer 3

You must have at least $3$ even numbers, and as many as $6$

How many combinations for each number of odd rolls?

$\sum_\limits{n=3}^{6} \frac {{8\choose n}}{2^8}$

Answer 4

You won't have at least 2 odds if there are exactly 7 or 8 evens. You won't have at least 3 evens if there are exactly 6, 7, or 8 odds.

The probability of 8 odds = $\frac{1}{2^8}$ and same for 8 evens.

The probability for 7 odds (or evens) = $\frac{1}{2^8}$ with 1 degree of freedom for the even roll to appear anywhere between the odds, or $8*\frac{1}{2^8}$.

The probability for 6 odds (with 2 evens) = $\frac{1}{2^8}$ with 2 degrees of freedom for the 2 evens to appear anywhere. The way I think about this intuitively is to imagine the 6 odds on the ground, picking up the 2 evens for the time being, and there being a bucket between and at the beginning and end of the 6 odds for me to place the 2 evens. So, there would be 7 buckets I can place my evens in, but I need to be careful not to double count my choices because the order of the evens does not matter. In the first scenario the 2 evens will go in the same bucket, and there are 7 ways that can happen. In the second scenario, they will go to different buckets and there are $\frac{7*6}{2}$ ways for that to happen (need to divide by two because choosing the 1st bucket then the 3rd is the same as choosing the 3rd bucket then the 1st). So, we get 28 from $7+21$ and this probability being $28*\frac{1}{2^8}$.

Side note: if you want to say for the bucket choices it is $7*7=49$, to remove the double counting you're essentially removing duplicates from the set {(1,1), (1,2), ..., (2,1), ..., (7,7)} which isn't as intuitive but would be $(49-7)/2+7$ (remove the same buckets, divide by 2, then add back in the same buckets).

So add them all up and take away from 1 to get:

$1 - 2*\frac{1}{2^8} - 2*8*\frac{1}{2^8} - 28*\frac{1}{2^8}$.

Also, 28 is also ${8}\choose{2}$ and most people will probably just automatically go with this format for this kind of problem but you can essentially derive it yourself if you just think about it.