<s>[INST] Confusing probability problems based on product rule and combinations I am going thru probability exercise. Faced first problem:

Book Q1. Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. What is the probability the largest number appearing on the selected tickets is 7?

My logic: if one of six tickets should be 7, the $\color{red}{\text{remaining 5}}$ can be any of 1 to 7, so it should be $7^5$. 
But turns out that the given solution is $\frac{7^6-6^6}{10^6}$. 
My Q1. Though I understood the logic behind $\frac{7^6-6^6}{10^6}$, I was wondering what is exact logical mistake with $7^5$? I guessed that $7^5$ completely ignores what should be 6th ticket, it only puts restriction on 5 tickets. Is it like that?
Then I came across similar but more involved problem, with significant difference from above one that it performs action without replacement:

Book Q2. Three numbers are chosen at random without replacement from (1,2,3,...,10). What is the probability that the minimum number is 3 or the maximum number is 7?

My logic: Noticing that this is without replacement, I guessed the solution should be 
$$
=
\begin{pmatrix} 
\text{selections with}\\ 
\text{minimum}\\ 
\text{number is 3} 
\end{pmatrix}
+
\begin{pmatrix} 
\text{selections with}\\ 
\text{maximum}\\ 
\text{number is 7} 
\end{pmatrix} 
-
\begin{pmatrix} 
\text{selections with}\\ 
\text{maximum}\\ 
\text{number is 7}\\ 
\text{and minimum}\\ 
\text{number is 3} 
\end{pmatrix} 
$$
$$=
\frac{
\overbrace{(\binom{8}{3}-\binom{7}{3})}^{\text{#selections with min 3}}
+
\overbrace{(\binom{7}{3}-\binom{6}{3})}^{\text{#selections with max 7}}
-
\overbrace{3\times {^3P_3}}^{\text{#selections with max 7 and min 3}}
}
{\binom{10}{3}
}
$$
But the book solutions says:

P(minimum 3) or P(maximum 7)
P(minimum 3) $=\frac{\binom{7}{2}}{\binom{10}{3}}=\frac{21}{120}$
P(max 7) $=\frac{\binom{6}{2}}{\binom{10}{2}}=\frac{15}{120}$
Thus the solution is $\frac{11}{40}$

My Q2. How even by books logic the solution $\frac{11}{40}$ is achieved. I am unable to understand it as I find the explanation insufficient.
My Q3. If book Q2 answer is correct then why for book Q1 solution is not $7^5$ which is what I initially guessed (because the only difference being with / without replacement, the logic of getting $\color{red}{\text{remaining m}}$ stuffs out n should remain same)?
My Q4. If we make first question without replacement, will the solution be $\frac{\binom{7}{6}-\binom{6}{6}}{\binom{10}{6}}$?
My Q5. What will be the solution if we make book Q2 with replacement?
My Q6. Where my logic for solution to Book's Q2 is wrong?
 [/INST] With the actual answers already given, I am putting different possible scenarios and there solutions here for reference.


*

*Selecting six out of {1,2,3,...,10} with replacement. Probability of largest selected number is 7 = ?
Sol. 
$$
\overbrace
{\frac{7^6-6^6}{10^6}}
^{
\begin{pmatrix}
\text{no. of ways of selecting any six out of {1,..,7}} \\
\text{removing those not containing 7}\\
\text{i.e. those six from {1,..,6}}\\
\end{pmatrix}
}=0.070993$$

*Selecting six out of {1,2,3,...,10} without replacement while taking order of selection into account. Probability of largest selected number is 7 = ?
Sol. 
$$
\frac{
\overbrace{\binom{7}{6}}
^{\begin{pmatrix}
\text{no. of ways to} \\
\text{select six from {1,..,7}} \\
\text{without considering} \\
\text{order} \\
\end{pmatrix}}
-
\overbrace{\binom{6}{6}}
^{\begin{pmatrix}
\text{no. of ways to} \\
\text{select six from {1,..,6}} \\
\text{without considering} \\
\text{order} \\
\end{pmatrix}}
}{
\binom{10}{6}
}=
\frac{
\overbrace{\binom{6}{5}}
^{\begin{pmatrix}
\text{selecting six from {1,..,7}} \\
\text{with 7 as max reduces to} \\
\text{selecting 5 from {1,..,6}} \\
\text{since you know you must} \\
\text{have 7} \\
\end{pmatrix}}
}{\binom{10}{6}}=\frac{1}{35}=0.028571
$$

*Selecting six out of {1,2,3,...,10} without replacement while not taking order of selection into account. Probability of largest selected number is 7 = ?
Sol.
$$
\frac
{\overbrace
{^6P_5}
^{
\begin{pmatrix}
\text{no. of ways to select five} \\
\text{out of {1,..,6} while} \\
\text{considering order} \\
\end{pmatrix}
}
\times
\overbrace
{6}
^{
\begin{pmatrix}
\text{putting 7} \\
\text{in one of 6 places} \\
\end{pmatrix}
}
}{^{10}P_6}
$$

*Selecting three out of {1,2,3,...,10} with replacement. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$\frac{(8^3-7^3)=(7^3-6^3)-18}{10^3}=0.278$$
For details, see correction in Brian Scott's solution.

*Selecting three out of {1,2,3,...,10} without replacement while taking order of selection into account. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$
\frac
{\overbrace
{^7P_2\times 3}
^{
\begin{pmatrix}
\text{no. of ways to select two} \\
\text{out of {4,..,10}}  and \\
\text{put 3 in any of three places} \\
\text{making 3 min number among} \\
\text{three selected} \\
\end{pmatrix}
}
+
\overbrace
{^6P_2\times 3}
^{
\begin{pmatrix}
\text{no. of ways to select two} \\
\text{out of {1,..,6} put 7}  and \\
\text{put 7 in any of three places} \\
\text{making 7 max number among} \\
\text{three selected} \\
\end{pmatrix}
}
-
\overbrace
{^3P_3 \times 3}
^{
\begin{pmatrix}
\text{Selecting 3 tickets with both} \\
\text{min 3 and max 7:} \\
\text{first select any one of} \\
\text{4,5 or 6,then perform} \\
^3P_3\text{ arrangements with the} \\
\text{selected number,3 & 7} \\
\end{pmatrix}
}
}{^{10}P_3}
= \frac{198}{720} = 0.275
$$

*Selecting three out of {1,2,3,...,10} without replacement while not taking order of selection into account. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$
\frac
{\overbrace
{\binom{8}{3}-\binom{7}{3}}
^{
\begin{pmatrix}
\text{no. of selections} \\
\text{with 3 as minimum} \\
\end{pmatrix}
}
+
\overbrace
{\binom{7}{3}-\binom{6}{3}}
^{
\begin{pmatrix}
\text{no. of selections} \\
\text{with 7 as maximum} \\
\end{pmatrix}
}
-
\overbrace
{3}
^{
\begin{pmatrix}
\text{no. of selections with both} \\
\text{3 as min & 7 as max:} \\
\text{{3,7,4},{3,7,5},{3,7,6}} \\
\end{pmatrix}
}
}{\binom{10}{3}}
=\frac{\binom{7}{3}+\binom{6}{2}-3}{\binom{6}{2}}
=\frac{11}{40}
=0.275
$$
</s>