Probability that a man will hit the target The question is


A man can hit a target once in $4$ shots. If he fires 4 shots in succession, what is the probability that he will hit his target? 


Here is how I am solving it:
Since the probability of man hitting the target is $\frac{1}{4}$ so for four consecutive shots it will be $(\frac{1}{4})^4 = \frac{1}{256}$ which is wrong.
Now the book takes a different approach and finds the probability that he will not hit the target in one shot = $1 - \frac{1}{4} = \frac{3}{4}$ therefor the probability he will not hit the target in 4 shots is  $(\frac{3}{4})^4$ and thus , the probability that he will hit the target at least in one of the four shots is 1- $(\frac{3}{4})^4$
Although I understand the books approach - I wanted to know why my approach is wrong ? doesnt it also calculate the probability of hitting the target in 4 shots
 A: Your first calculation finds the probability that the person hits the target $4$ times in a row. That is very different (and much smaller) than the probability that the person hits at least once.
Let's do the problem in another way, much too long, but it will tell us what is going on. 
What is the probability of hitting at least once? The event "at least one hit" could happen in several ways: (i) exactly $1$ hit; (ii) exactly $2$ hits; (iii) exactly $3$ hits; (iv) exactly $4$ hits.
(i) The probability of exactly one hit is $\binom{4}{1}(1/4)(3/4)^3$. This is because the hit could happen in any one of $4$ (that is, $\binom{4}{1}$) places. Write H for hit and M for miss. The probability of the pattern HMMM is $(1/4)(3/4)(3/4)(3/4)$. Similarly, the probability of MHMM is $(3/4)(1/4)(3/4)(3/4)$. You will notice this probability is the same as the probability of HMMM. We get the same probability for MMHM and for MMMH, for our total of $\binom{4}{1}(1/4)(3/4)^3$.
(ii) Similarly, the probability of exactly $2$ hits is $\binom{4}{2}(1/4)^2(3/4)^2$. 
(iii) The probability of $3$ hits is $\binom{4}{3}(1/4)^3(3/4)$.
(iv) The probability of $4$ hits is $\binom{4}{4}(1/4)^4$. This is the $(1/4)^4$ that you calculated.
Add up. We get the required answer. 
However, that approach is a lot of work. It is much easier to find the probability of no hits, which is the probability of getting MMMM. This is $(3/4)^4$. So the probability that the event "at least one hit" doesn't happen is $(3/4)^4$. So the probability that the event "at least one hit" does happen is $1-(3/4)^4$. 
A: This is a very Basic Probability Question. These methods are Verified although I don't know who is Downvoting it every time.
Please read carefully to understand.
Formulae to be Used :
Sum of Nth terms of G.P : $S_n = a\frac{(1 – r^{n})}{(1 – r )}$
Let's say, he fires at the target K times. Then, He can Kill his enemy by his 1st Hit, 2nd Hit, 3rd Hit, and so on.
$P(e)$ =>  Man Successfully HITS the target
$P(\bar{e})$ =>  Man Failing to Hit the target
Succcess OR Fail Succcess OR Fail Fail Succcess OR... and so on  can be written as :   $P(e)  + P(\bar{e})P(e) + P(\bar{e})P(\bar{e})P(e) + .... $ and so on
Therefore, P(Hitting the Target At least once) =
$ P(e) + P(\bar{e})P(e) + P(\bar{e})P(\bar{e})P(e) + ..... [P(\bar{e})....(K-1) times ]P(e)  = P(e) [ 1 + P(\bar{e}) +  P(\bar{e})^{2} +  P(\bar{e})^{3} + ... +  P(\bar{e})^{K - 1 } ]  = [\frac{1 - P(\bar{e})^{K}}{1 -  P(\bar{e})}]P(e) = [\frac{1 - P(\bar{e})^{K}}{P(e)}]P(e) = [1 - P(\bar{e})^{K} )] = [1 - (1 - 1/4)^{4}] = 1 - (3/4)^{4} $
Let's take another question :
Question 2 : What is the probability of his Hitting the Target at least Twice ?
Solution of Question 2:
$n = 4$
$ Success = p  = 1/4$
$Failure = F = q = 1- 1/4  = 3/4$
We will use the formulae : $\binom{n}{r}p^{r}q^{n-r}$ and subtract 1 from it.
$P(F≥2) = 1 −[P(F= 0) + P(F= 1)] = 1 - [\binom{4}{0}.(\frac{1}{4})^{0}.(\frac{3}{4})^{4} + \binom{4}{1}.(\frac{1}{4})^{1}.(\frac{3}{4})^{4-1} ]=1-[\binom{4}{0}.(\frac{1}{4})^{0}.(\frac{3}{4})^{4} + \binom{4}{1}.(\frac{1}{4})^{1}.(\frac{3}{4})^{3} ] $
