understanding of replacement and without replacement let us consider following problem  : 
A Harris poll found that $46$% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.
of course standard approach will be  following :
let S assume event  that selected person suffer great stress, then we will have
$P(SSS)=0.46*0.46*0.46=0.097336$
but let us ask  question like this : i went into street and selected first person, then second  then third, if probability of  first person who suffers stress is $0.46$ and if i  selected such person, then for the second one it will be  different right? let us consider following simplification, we have $100$ person,  $0.46$ probability means that out of $100$, we have $46$ we  suffer from stress, now i  have selected one of them  , so totally left $99$ person, among them there are $45$ who suffers stress and  therefore probability will be division of  $99$ and $45$ or $0.454545455$ , for the  third person  it will be division of  $44$ and  $98$ or  $0.448979592$
if we follow to this way, we will get
$P(SSS)=0.46*0.454545455*0.448979592=0.093877551$ 
so  how can i determine  when i should use conditional probability? should it be indicated in task?
 A: Your first answer $(.46)^3$ is reasonable in this problem. The number $X$ of
people who suffer stress in a sample of size $n = 3$ has
$S \sim \mathsf{Binom}(n=3, p = .46).$ You have found $P(X=3) = .46^3 = 0.0973.$
It is easy to find $P(X = 0) = 0.1575,\; P(X = 1) = 0.4024$ and 
$P(X = 2) = 0.3428$ to get the entire PDF. (I have rounded to four places
which is much more accurate than you would need in practice.)
A distribution table from R statistical software is as follows:
x = 0:3;  b.pdf=dbinom(x, 3, .46)
cbind(x, b.pdf)
     x    b.pdf
##   0 0.157464
##   1 0.402408
##   2 0.342792
##   3 0.097336

A general rule is that one does not try to distinguish between sampling
with replacement (binomial) and sampling without replacement (hypergeometric)
unless one is sampling more than 10% of the population.
To illustrate this, let's imagine there are only 1000 people in the population,
of which 460 suffer stress, and the remaining 540 do not. (Of course, there 
really are many more than 1000 people, but that is a big enough number
for my illustration.)  Then the hypergeometric distribution, which keeps
track not to ask anyone more than once, is given in the second column below.
The third column shows the binomial probabilties from above. You can see that
the probabilities are very nearly the same.
x = 0:3;  h.pdf=dhyper(x, 460, 540, 3)
cbind(x, h.pdf, b.pdf)
 x     h.pdf    b.pdf
## 0 0.1570611 0.157464
## 1 0.4028706 0.402408
## 2 0.3430753 0.342792
## 3 0.0969929 0.097336

Another issue to consider is that the Harris poll has a margin of error
in its findings. Perhaps what is quoted as $46\% = 0.46$ is really $0.46 \pm 0.02,$ which would be the case if Harris interviewed about 2500 people about
stress. Then your original answer $.46^3=0.0973$ could have been as
small as $.44^3 = 0.0852$ or as large as $0.111.$
The error from ignoring sampling with replacement is insignificant compared
to the error in the original number 0.46. (Maybe this is one of the 'other
problems' @lulu had in mind.)
