Difficulty in understanding the probability exactly one type i in the final collection out of n types It is Question 63 on page 174 in Ross's book (Introduction to Probability Models-11th edition)
Suppose that there are n types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of n types. Suppose one continues collecting until a complete set of at least one of each type is obtained. 
(a) Find the probability that there is exactly one type i coupon in the final collection.
Hint: Condition on T, the number of types that are collected before the first type ii appears.
The same question can be found here, but I do not have enough credit to comment on.
A conditional probability problem on coupon collection
The Solution Says:

Let Si be the event there is only one type i in the ﬁnal set.

P{Si = 1}
= $\sum_{j=0}^{n-1}$ P{Si=1|T=j}P{T=j}
= $1/n$ $\sum_{j=0}^{n-1}$ P{Si=1|T=j}
= $1/n$ $\sum_{j=0}^{n-1}$ $1/(n-j)$
The ﬁnal equality follows because given that there are still n−j−1
uncollected types when the ﬁrst type i is obtained, the probability starting
at that point that it will be the last of the set of n−j types consisting of
type i along with the n−j−1 yet uncollected types to be obtained is, by 
symmetry, 1/(n−j).

Here are my questions:


*

*Why $\sum_{j=0}^{n-1}$ P{Si =1|T=j}P{T=j} = $\sum_{j=0}^{n-1}$P{Si=1|T=j}?   

*How to understand P{Si=1|T=j} = $1/(n-j)$ ? Especially when $j = 0$, it is $1/n$? Does that mean the whole branch starting from coupon i is $1/n$? If it is the case, I think case like the instance i-i-... is invalid.
Your help will be greatly appreciated.
 A: For part 1 of the question,
$$ \sum_{j=0}^{n-1} P\{S_i =1\mid T=j\}P\{T=j\} \neq
 \sum_{j=0}^{n-1} P\{S_i=1\mid T=j\},$$
and the solution never claimed that those two quantities are equal.
What the solution says is that $P\{T=j\} = \frac1n,$
and therefore
\begin{align}
\sum_{j=0}^{n-1} P\{S_i =1\mid T=j\}P\{T=j\}
&= \sum_{j=0}^{n-1} P\{S_i =1\mid T=j\}\times\frac1n \\
&= \frac1n\sum_{j=0}^{n-1} P\{S_i =1\mid T=j\}.
\end{align}
You omitted the factor $\frac1n$.
For part 2, let's consider the case $j=0.$
In that case we want to evaluate  $P\{S_i = 1 \mid T=0\}.$
The event $T=0$ means that the collector has received no coupons of any other type at the time they receive their first coupon of type $i.$
The only way for this to happen is if the very first coupon received
is a coupon of type $i.$
So now we have established that we are looking for the probability that
$S_i = 1,$ given that the very first coupon is of type $i.$
The "given" means we can consider "first coupon is of type $i$" as an event
that has already happened.
We will then have $S_i = 1$ if and only if the following chain of events happens:


*

*Starting from time $t_i$ (the moment after receiving the first coupon), the collector continues to collect coupons;

*After time $t_i,$ the collector collects at least one of each of the other $n - 1$ types of coupons;

*Between time $t_i$ and the time when the collector collects the last of the other $n - 1$ types of coupons, the collector does not receive any coupon of type $i.$


Let's call this chain of events $A_1.$
Then $P\{S_i = 1 \mid T=0\} = P\{A_1\mid T=0\}.$
Now let's compare that chain of events to the following chain of events
that occur to a different collector who starts collecting coupons at time $t_i$ (that is, at time $t_i$ the second collector has no coupons at all):


*

*Starting from time $t_i,$ the collector collects coupons;

*After time $t_i,$ the collector collects at least one of each of the $n - 1$ types of coupons that are not of type $i$;

*Between time $t_i$ and the time when the collector collects the last of the other $n - 1$ types of coupons, the collector does not receive any coupon of type $i.$


Let's call this chain of events $A_2.$
If the second collector also is motivated to collect all $n$ types of coupons, they will continue to collect coupons until they have received a coupon of type $i,$ which will certainly happen eventually;
and then type $i$ will be the last type of coupon that collector collects.
So $P\{A_2\mid T=0\}$ is exactly the probability type $i$ will be the last type of coupon collected by a collector who collects all $n$ types of coupons (starting at time $t_i$), given that $T=0.$
Now, the chance that type $i$ will be the last type you collect is independent of when you start collecting, so we can ignore the part about
"starting at time $t_i.$"
The given fact that $T = 0$ likewise has no effect on the chances of the second collector to collect coupons; it is merely something that happened to the first collector.
So we're down to just, "What is the chance that type $i$ is the last of the $n$ types collected?"
And that probability is $\frac1n$ by symmetry; that is, 
$P\{A_2\mid T=0\} = \frac1n.$
But now compare the chain of events $A_1$ to the chain of events $A_2.$
In both events we have a collector collecting all types of coupons except type $i$ between time $t_i$ and some later time.
There is nothing to distinguish between the probabilities of those two events; $P\{A_1\mid T=0\} = P\{A_2\mid T=0\}.$
Therefore $P\{A_1\mid T=0\} = \frac1n,$
and therefore $P\{S_i = 1 \mid T=0\} = \frac1n.$
For $j > 0$ there is an additional subtlety: after $t_i$ (when the collector gets the first coupon of type $i$), we ignore the existence of the $j$ types of coupons collected before $t_i.$
With regard to the question of the collector collects all the other 
$n - j - 1$ types of coupon before getting another coupon of type $i$
after time $t_i,$
the receipt of one of those $j$ types coupons has as much impact as receiving a postcard from Aunt Sally.
So we treat the problem as one involving only $n - j$ types of coupon,
namely type $i$ and the $n - j - 1$ types not yet collected,
and we want the probability of collecting all of the other
$n - j - 1$ types before collecting (the next) type $i.$
That probability is $\frac1{n-j}$ for the same reasons as in
the case $j = 0.$
