Simultaneous occurrence of two events I am facing a serious conceptual problem.
Suppose a bag has 4 white balls and 5 red balls.
If two balls are drawn one by one, then what is the probability of both balls being red? (without/with replacement)
Is this a example of of two events occurring simultaneously?
But if a ball is drawn first and next ball is drawn after it, how they can be both red at the same time?
(As the one ball is drawn first, it can be red before the second ball is drawn.)
How can both balls can be red at the same time ?
(I have no problem to understand that both balls can be red, but it can't happen simultaneously as both balls are not drawn at the same time.)
 A: 

If two balls are drawn one by one, then what is the probability of both balls being red?(without/with replacement)

Is this a example of of two events occurring simultaneously?

For the above experiment,let event $A$ be the first ball being red, i.e., $A=\{RR,RW\};$
let event $B$ be the second ball being red, i.e., $B=\{RR,WR\};$
let event $C$ be both balls being red, i.e., $C=A\cap B=\{RR\}.$
The sample space  of a probability experiment is its set of outcomes, each outcome giving the result of all its trials (here: ball draws). An event is some subset of the sample space, so is a specific collection of outcomes. Remember, an event is not generally a single experiment outcome (such a single-outcome event is called an elementary event) nor a trial result ($A$ and $B$ are examples of such an event); rather, think of it as a particular meta-level ‘snapshot’ of outcomes. So, an experiment's events don't have an inherent sequence. (After all, formulations and rules in classical logic—so, mathematical reasoning and probability—are agnostic to time/tense.)
Thus, events $A$ and $B$ simultaneously occurring makes sense, and is precisely event $C$ occurring. The event of the first ball being red (i.e., event $A=\{RR,RW\})$ is the set of outcomes such that the first ball is red, rather than the act/incident of drawing the first ball. We might also have an event $F$ of obtaining no red ball, which, again, is neither a particular experiment outcome nor a trial result but a particular snapshot of the experiment.
In fact, we can reframe the probability experiment (and obtain all the same results) by considering the second draw as Trial 1 and the first draw as Trial 2 even when the trials are not independent; I did just that here, to simplify the analysis. Furthermore, if the draws are independent, then successive draws (from the same bag) can be equivalently regarded as simultaneously draws (from different bags), in other words, we can just treat ‘second draw’ as ‘Bag B (which is drawn from simultaneously as Bag A’).
A: 
Is this a example of of two events occurring simultaneously?

No. It's "one by one" not "together/simultaneously". So, one ball is drawn first then depending upon the question it is replaced or not replaced into the box and then the next ball is drawn.
We can have $2$ balls drawn such that both are red. Better/correct word to use in this case would be "consecutively".
Now follow the below reasoning:

What if we were told that they were drawn simultaneously?
Let's say we are given that 2 balls are drawn simultaneously. What is the probability that both the balls are red?

If I'm picking $2$ balls simultaneously, then the $2$ balls can be simultaneously red too. We would solve it this way:
$M1:$ Long way
$E=\{(R_1R_2),(R_1R_3),(R_1R_4),(R_1R_5),(R_2R_3),(R_2R_4),(R_2R_5),(R_3R_4),(R_3R_5),(R_4R_5)\}$
$S=\{(W_1R_1),(W_1R_2),\dots,(R_3R_5),(R_4R_5)\}$
$n(E)=10, n(S)=36$
$P(E)=\frac{n(E)}{n(S)}=\frac{10}{36}=\frac5{18}$
$M2:$ Shorter way
$n(E)=$ $^5C_2=10$
$n(S)=$ $^9C_2=36$
$P(E)=\frac{n(E)}{n(S)}=\frac{10}{36}=\frac5{18}$
Note 1: "Simultaneously" means happening at same time. So there's no way one of the 2 balls, that is, the $1^{st}$ ball can be ever replaced. So, it would automatically mean that the replacement can't happen. That's why the $^nC_2$ worked.
Hope you can now work out yourself cases such as: $3$ dice are thrown smly, $4$ cards are picked smly from a standard deck, etc. where smly means simultaneously.
In some cases, it doesn't matter whether the experiment is one by one or simultaneously. Like that of rolling of multiple dice or tossing of multiple coins, etc. here the sample space will the same. (Basically those cases where replacing anyways can't happen).
