Probability of throwing missiles "all at once" and "one by one" 5 missiles can be thrown at a target with probability of hitting the target for each missile = 0.3 .It is done in two ways :


*

*One by one assuming that no more missiles are thrown if the target is hit so there is no need of including the probability of not hitting the targets of other missiles in the calculation . So the binomial distribution method won't work. 


Since no more missiles are thrown if the target is hit the binomial distribution method won't work.
Then the probability of hitting the target "only" once can be calculated as follows :
(Probability of first missile hitting the target) + (Probability of first missile missing the target) * (Probability of second missile hitting the target) + (Probability of first missile missing the target) * (Probability of second missile missing the target) * (Probability of third missile hitting the target).........and so on till the 5th missile .
=> 0.3 + (1-0.3)0.3 + (1-0.3)(1-0.3)*0.3 +........
=> 0.3 + 0.7*0.3 + 0.7*0.7*0.3 +........
Taking 0.3 common in the above expression,
0.3*[1 + 0.7 + (0.7)^2 + (0.7)^3 ......]
The expression in the square brackets can be solved by the geometric sum formula as,
=> 0.3*[1*(1- 0.7^5)]/0.3
= 1-0.7^5 = 0.83193


*All at once all 5 of them


Then the probability of hitting the target "at least" once can be calculated  as:
The probability of hitting the target at least once = 1-(Probability of all the missiles missing the target) 
= 1 - (Probability of one missile missing the target)^5 
= 1 - (1-0.3)^5 = 1 - (0.7)^5 = 0.83193 
Which is same as the first case .
Its really interesting that the probability of hitting target only once if we throw missiles one by one is same as hitting it at least once if we throw them all at once .
My question is why is this happening ? What is the reason behind it ? Is it just a coincidence ? (I know the question doesn't sound much mathematical and I am sorry if this violates the posting policy ; if it does , then I will sincerely take it down) .
 A: The first scenario (throwing missiles until you either stop or hit 5 throws) can be modeled as following a geometric distribution, where we are only calculating up to $n=5$:
$$P(\text{Success})= 0.3\sum_{k=1}^5 0.7^{k-1} = 0.83193$$
For throwing all 5 at once, you are performing a binomial experiment:
Let $X$ be the number of missiles that hit the target. We succeed as long as they don't all miss.
$$P(X>0)=1-P(X=0)=1-0.7^5 = 0.83193$$
Same probabilities, as you calculated. So your question is why probabilities calculated for a stopped sequence is the same as for a simultaneous group.
At a non-technical level, we can imagine that you continue to throw missiles after your first hit and then aggregate all sequences that are successful. This will reproduce the same set of outcomes as the binomial case where you throw them all at once. Now, note that all possible outcomes after the first hit are counted as part of the probability of the first hit (i.e., the probability that you hit it on your first throw are higher than hitting it on your second, because it counts more of these events after the first hit). 
This really just means that the probability of hitting on your second throw has the same probability as the sum of all binomial trials whose first hit is at the second "position" (assuming you numbered your missiles in increasing order). 
A: The reason why the two figures will always be equal is very simple.
One by one
We have "won" as soon as a hit is made, and stop. $(H =\; hit,\;\; M = \;miss)$
Suppose we hit on the third trial: 
$M-M-H-?-?$
Trials $4$ and $5$ won't take place, but what happens if they took place doesn't matter,
we have won anyway, so we can as well include all possibilities after the third trial, viz:
$M-M-H||-H-H$
$M-M-H||-H-M$
$M-M-H||-M-H$
$M-M-H||-M-M$
Including all possibilities after the third means we are multiplying $\;\Bbb P(M-M-H)$ by $1$,
thus $\;\;\Bbb P$(stopping after the first hit) =  $\Bbb P$(getting at least $1$ hit)
