Probability of Finding Oil 
An oil prospector will drill a succession of holes in a given area until he finds r productive wells. The probability that he is successful on a given trial is 0.3. The prospector can only afford to drill 6 wells. What is the probability the inspector fails to find r productive wells if:
  (a) r = 1? (b) r = 2? (c) r = 3?

 I believe I did the work correctly I just want some clarification.
Let S=sucuess and F=Failure.  So $P(S)=.3$ and $P(F)=.7$.  So to find each we would have to do a geometric disturbution, $p(r)=P(F)^{6-r}P(S)$
 So for a, we have: $p(r=1)=(.7)^{6-1}(.3)=.050421$
 For b: $p(r=2)=(.7)^{6-2}(.3)=.07203$
 And for c: $p(r=3)=(.7)^{6-3}(.3)=.1029$

The probability of success for a given $r$ is the probability to have at least $r$ successes when attempting $6$ times an expriment with a probability of success equal to $p=0.3$.
That is, if $X$ is random variable following a binomial ditribution $B(6,0.3)$, the probability $p_r$ of success, for a given $r$, is
$$p_r=p(X\ge r)$$
Since $X\ge r+1\implies X\ge r$, this also means that $p_{r+1}\le p_{r}$, that is $p_r$ is decreasing as $r$ increases.
Now,
$$p_1=p(X\ge1)=1-p(X=0)=1-(1-p)^6\simeq0.882351$$
$$p_2=p(X\ge2)=1-p(X\le1)=1-(1-p)^6-\binom{6}{1}p(1-p)^5\simeq0.579825$$
$$p_3=p(X\ge3)=1-p(X\le2)=1-(1-p)^6-\binom{6}{1}p(1-p)^5-\binom{6}{2}p^2(1-p)^4\simeq0.25569$$
It may look like we are cheating, since the guy is supposed to stop when he reaches $r$ successes. But we can pretend he then continues to drill until he reaches $6$, as this won't change the outcome. That's why we must check if $X$ is at least $r$.
