One way to check theory is to simulate the problem. The code snippet below (using Octave) gives the empirical value of $0.62980$ which is comparable to the answer you have: $0.62419$ (See computation at Wolfram Alpha).
Of course, simulations like the one below are no guarantee that your theoretical answer is correct. However, at the very least they can help rule out incorrect answers and serve as a secondary check to your theoretical analysis.
function ev = ProbEst()
# Probability of hit is 1/5
probHit = 1/5;
totalIter = 10000;
noShots = 10;
# Counter to track if the event of interest
# has occurred
eventOccurred = 0;
# Set seed of random number generator
rand("state",[11;22;3209;42038021;11]);
for iter =1:totalIter
targetHit = 0;
for shot = 1:noShots
if (rand(1,1) < probHit)
targetHit += 1;
endif
endfor
if (targetHit >= 2)
eventOccurred += 1;
endif
endfor
# Output empirical probability to console
eventOccurred/totalIter
endfunction
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 $1-(3/4)^4$.
Best Answer
Hints: For ii, you have the probability of missing every time from i, so all the rest is the chance of hitting at least once.
For b, the same logic of deriving ii from i applies, but the number of shots is a variable. If you want 0.95 to hit at least once, what is the chance you miss them all? Then find a number of shots that gets the chance below that.