In which order do I use OR in this probability question?

Question

I'm given the question:

A dice is rolled $15$ times. What is the probability that either a $1$ or $6$ is scored exactly eight times?

Now I know that the probability $1$ is scored exactly $8$ times is...

$Let \space P({X_n}=m)= Probability \space n \space is \space scored \space m \space times$

$P({X_1}=8)={15\choose 8} ({1\over 6})^8 ({5\over 6})^7$

And since it's a fair dice, likewise...

$P({X_6}=8)={15\choose 8} ({1\over 6})^8 ({5\over 6})^7$

Now about the "either $1$ or $6$" part, do I do that FIRST as in...

$P({X_1} \cup {X_6}) = {2\over 6}$

Then do...

$ {15\choose 8} ({2\over 6})^8 ({4\over 6})^7 $

OR...

Is it meant to be done as the final step like so...

$P( P({X_1}=8) \cup P({X_6}=8))$

I hope my question makes sense.

Answer 1

By inclusion-exclusion, \begin{align*} P(X_1=8\cup X_6=8) &= P(X_1 = 8)+P(X_6 = 8)-P(X_1 = 8, X_6 = 8)\\ & = \binom{15}{8}\left(\frac{1}{6}\right)^8\left(\frac{5}{6}\right)^7+\binom{15}{8}\left(\frac{1}{6}\right)^8\left(\frac{5}{6}\right)^7+0\\ &= 2\binom{15}{8}\left(\frac{1}{6}\right)^8\left(\frac{5}{6}\right)^7 \end{align*} where $P(X_1 = 8, X_6 = 8)=0$ since you can't roll 8 ones and 8 sixes since you only roll 15 times.

---

A dice is rolled $15$ times. What is the probability that either a $1$ or $6$ is scored exactly eight times?

If the interpretation is that on each roll, you treat a one and a six a success (you are interested in rolling a one or a six), then let $$Y_i = \{\text{Roll a six or a one on roll $i$}\}.$$ Then let $S = Y_1+\dotsc+Y_{15}$ which follows $\text{Bin}(15, p = 2/6)$ and so $$P(S = 8) = \binom{15}{8}\left(\frac{2}{6}\right)^8\left(\frac{4}{6}\right)^7.$$

Answer 2

To answer Arthur's interpretation of probablyme's answer, note that the probability of rolling either a $1$ or a $6$ on a given roll is $1/3$, so the probability of obtaining either a $1$ or a $6$ on exactly $8$ of the $15$ rolls is

$$ P = \binom{15}{8}\left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^7 $$

My advice is to determine, somehow, which interpretation of the question is intended, and answer accordingly. Failing that, specify both interpretations and answer both.

Answer 3

The question is a little ambiguous.   Your calculations for the possible interpretations are correct.   However, your expressions of what you are calculating is a little off.

Let $X_n$ be the count of occurrences of result $n$ in $15$ rolls of a fair die.   Then we are required to find one of these probabilities:

$$\begin{align}\mathsf P(X_1=8\cup X_6=8) \; = \;& 2\;\dbinom{15}{8}\dfrac{5^7}{6^{15}} & \textsf{either obtain 1, or 6, exactly eight times } \\[2ex] \mathsf P(X_1+X_6=8) \; = \; & \dbinom{15}{8}\dfrac{2^8\,4^7}{6^{15}} & \textsf{obtain, either 1 or 6, exactly eight times}\end{align}$$