A die is rolled 6 times; what is the chance that the first roll is a one or the last roll is a one. This question appears on page 244 of "Statistics, 4th ed" by David Freedman.
The text of the question is:
A die is rolled 6 times. 
The chance that the first roll is an ace or the last roll is an ace equals ______. 
Answer choices are: $(1/6 + 1/6)$, $(1/6 * 1/6)$, and "neither of these".
The correct answer, according to the book is "neither of these".
I tried calculating the probability assuming that any number of aces can appear in the six rolls but that the first and the sixth rolls must result in aces.
I got ~0.278 as the answer. 
$(1/6)^2*(5/6)^4 + (1/6)^3*(5/6)^3*4!/3! + (1/6)^4*(5/6)^2 * 4! / (2!*2!) + (1/6)^5*(5/6)*4!/3! + (1/6)^6$
Is this correct?
My question is: how can one infer from this question that we are to consider all cases of an "ace" appearing on any of the six rolls. Initially, I thought I could consider the six rolls as independent.
 A: You are correct that the six rolls are independent. So the probability that the first roll is an ace is $1/6$, and similarly, the probability that the last roll is an ace is $1/6$. However, the probability of either event happening is not just the sum: you need to subtract the probability of their intersection. The probability of both the first and last roll being aces is $1/6^2$, so the final answer is $1/6 + 1/6 - 1/6^2 = 11/36$.
A: The probability that neither the first or last die is an ace is $(5/6)(5/6) = 25/36$, so the complementary event has probability $1 - 25/36 = 11/36$.
A: I just wanted to present some notation for practice, in case you are not familiar. Let $A = \{\text{First roll is one}\}, B = \{\text{Last roll is one}\}$. Then, recalling the inclusion-exclusion rule,
\begin{align*}
P(A\text{ OR }B) &= P(A)+P(B)-P(AB)\\
&=P(A)+P(B)-P(A)P(B) \\
&= \frac{1}{6}+\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}\\
&=\frac{11}{36},
\end{align*}
where the $P(AB)$ is the probability of $A$ and $B$ occurring and it  can be expressed as the product $P(A)P(B)$ since the events are independent. Alternatively, 
$$P(A\text{ OR }B) = 1 -P(A^cB^c) = 1-P(A^c)P(B^c) = 1-\frac{5}{6}\cdot\frac{5}{6} = \frac{11}{36},$$
by the complement rule.
Final note, if $A$ and $B$ were disjoint (mutually exclusive), then $P(AB) = 0$
and 
$$P(A\text{ OR }B) = P(A)+P(B)-P(AB) = P(A)+P(B)-0 = P(A)+P(B).$$
But this is not the case.
A: The probability that the first roll is a 1 is 1/6.  The probability that the last roll is a 1 is 1/6.  Since you're looking at the first OR the last role is a 1, you add the two probabilities.  So the answer is 1/6 + 1/6 = 2/6 = 1/3.
