Question I think may contain Conditional Probability I am currently studying for an exam and passed by this question. I am not really sure how to solve it, i feel like I should be using conditional probability but I feel like I'm not doing it right. 
This is the question.
A batter is workout out a new pitching machine and has a 70% chance of hitting the ball
a) find the probablity that the batter hits the first ball
b)find the probabality that the batter misses the first ball but hits the second ball
c)find the probablity that the batter misses the first two balls but hits the third ball
I forgot to paste choice d) which is d)find the probbality that the batter takes three or more swings before he hits a ball. I did (3/10)^3 = .027 or 2.7%. Is this correct?
 A: Your question indicates to use Geometric Distribution:
If '$p$' is probability of success, then probability of first success before '$k$' failures: $ (1-p)^{k}*p $
A: I get:
a: $70\%$, b: $21\%$, c: $6.3\%$,
using $\left(\dfrac{3}{10}\right)^m\dfrac{7}{10}$, where $m=0,1,2$ respectively.
d is:
$$\frac{7}{10}\sum_\limits{i=3}^\infty \left(\frac{3}{10}\right)^i$$
$$=\frac{7}{10} \left(\frac{3}{10}\right)^3\sum_\limits{i=0}^\infty \left(\frac{3}{10}\right)^i$$
$$=\frac{7}{10} \left(\frac{3}{10}\right)^3\frac{1}{1-\frac{3}{10}}$$
$$=\frac{7}{10} \left(\frac{3}{10}\right)^3\frac{10}{7}$$
$$=\left(\frac{3}{10}\right)^3$$
$$=2.7\%$$
A: These are independent events. i.e. The probability that he hits the second ball does not depend on whether or not he hits the first ball.  Therefore we do not need to rely on conditional probability.
a) The probability that he hits the first ball is $\frac{7}{10}$
b) The probability that he misses the first ball but hits the second ball is $\frac{3}{10}\times\frac{7}{10}$ which equals $\frac{21}{100}$
c) The probability that he misses the first two balls but hits the third ball is $\frac{3}{10}\times\frac{3}{10}\times\frac{7}{10}$ which equals $\frac{63}{1000}$
