Expectancy of 10 mutually independent random variables and conditional probability I've been given the following question, which has 2 parts:

Consider 10 (mutually) independent experiments (each is a random variable) numbered 1 through 10. The probability that the i-th experiment will be over within an hour from the moment it began is $\frac{i}{10}, 1\leq i\leq 10$

1st part:

(Assuming all the experiments are started at the same time,) what is the
expected number of experiments that will be over within an hour?

The solution just says 1 (without further explanation) but I disagree. At least one (the 10th experiment) will be over within an hour with probability 1, and the rest also have non-zero probability to end within the same time, so the expectancy has to be at least greater than 1.
My way of solving this was to define each experiment as a random variable $x_i$ such that
$$
x_i=
\begin{cases}
\displaystyle 1,  & \text{w.p. $\frac{i}{10}$ } \\
0, & \text{w.p. $\frac{10-i}{10}$ }
\end{cases}
$$
each with expectancy $E[x_i]=\frac{i}{10}$,
and then define $$Y=\sum_{i=1}^{10} x_i$$
Thus, the expectancy would be
$$
E[Y]=E[\sum_{i=1}^{10} x_i]=\sum_{i=1}^{10}E[x_i]=\sum_{i=1}^{10}\frac{i}{10}=5.5
$$
By linearity of expectation.
To verify my solution, I tried calculating $P(y)$ for each $y, 0\leq y\leq 10$ and then finding out the expectancy using its definition for discrete random variables. I've got the values for $y=0, 1, 9, 10$ but the rest were too difficult to generalize.
The 2nd part is as follows (it's unrelated to the 1st one, just to the main question):

What's the probability that the 5th experiment was over (within an hour), given a randomly selected experiment is over?

It wasn't quite clear in the question, but I'm assuming that they meant all the experiments had started, and after an hour we picked one randomly and it was over, what's the probability that it's the 5th one? I calculated the probability for a randomly selected experiment to be over by multipling the probability for each experiment to be over with the probability for choosing it (using the law of total probability). Each experiment is equally likely to be choosen, so
$$
P(B)=\sum_{i=1}^{10}\frac{i}{10}\frac{1}{10}=0.55
$$
Then, if $A_5$ is the event that the 5th experiment was over,
$$
P(A_5 | B)=\frac{P(B|A_5)P(A_5)}{P(B)}=\frac{1\cdot \frac{1}{10}}{0.55}=0.182
$$
But in the solution they defined $A_5$ as the event that the 5th experiment was choosen (so in general $P(A_i)=\frac{1}{10}$ for every $i$) and said
$$
P(A_5 | B)=\frac{P(B|A_5)P(A_5)}{P(B)}=\frac{\frac{5}{10}\cdot \frac{1}{10}}{0.55}=0.091
$$
But this doesn't make much sense to me. Did I interrupt the question differently than I should have?
Edit: Added the second part which wasn't included when I posted the question at first.
 A: Your calculations are correct.
If, for $1\leq i\leq 10$, the probability that experiment $i$ is over within an hour is $i/10$ then the expected count for such will be $5.5$ .
However, if, for $1\leq i\leq 10$, the probability that experiment $i$ is over within an hour is $1/10$ then the expected count for such will be $1$ .

So someone mixed up an $i$ with a $1$.

For your second question, $\mathsf P(B\mid A_5)$ is the probability that a randomly selected experiment is over, given that the fifth experiment is over.  Since the experiments are independent, then under that condition each will have the unconditional probability of being over, except the fifth.  Since under the condition the fifth has certainly ended.
Therefore we should have:
$$\begin{align}\mathsf P(B\mid A_5) &=\dfrac1{10}\left(\sum_{i=1}^4\mathsf P(A_i\mid A_5)+\mathsf P(A_5\mid A_5)+\sum_{i=6}^{10}\mathsf P(A_i\mid A_5)\right)\\&=\dfrac1{10}\left(\sum_{i=1}^4\dfrac i{10}+1+\sum_{i=6}^{10}\dfrac i{10}\right)\\&=\\&=\dfrac 1{10}\left(\dfrac{1+2+3+4}{10}+1+\dfrac{6+7+8+9+10}{10}\right)\\&=\dfrac{6}{10}\\\mathsf P(A_5\mid B)&=\dfrac{\dfrac 6{10}\cdot \dfrac{5}{10}}{\dfrac{55}{100}}\\&=\dfrac{6}{11}\end{align}$$
Or if the probability was $1/10$ for all:
$$\small\begin{align}\mathsf P(B\mid A_5) &=\dfrac 1{10}\left(\dfrac{4}{10}+1+\dfrac{5}{10}\right)\\&=\dfrac{19}{100}\\\mathsf P(A_5\mid B)&=\dfrac{\dfrac {19}{100}\cdot \dfrac{1}{10}}{\dfrac{1}{10}}\\&=\dfrac{19}{100}\end{align}$$
