How to solve Bernoulli process You are flipping a coin that lands on “heads” with probability 0.7. If you get the first tails on the kth flip, the probability that k is an integer multiple of 3 canbe expressed as a/b where the fraction is in its simplest form. Find a + b.
I am quite unsure on how to approach this problem. Considering the fact that you get a tail on kth flip, I can safely say that it's geometric i.e. (0.7)^(k-1) * (0.3)^k. If k is a multiple of three then k could be 3,6,9....infinity. Adding all these probabilities up, upto infinity would give me n * (a/b). But I'm not sure whether I am headed in the right direction or not. 
 A: to do it with Geometric Series:
As the probability of getting the first $T$ on trial $k$ is $.7^{k-1}\times .3$ we see that the desired result is $$\frac 3{10}\times \sum_{n=1}^{\infty} \left(\frac 7{10}\right)^{3n-1}=\frac 3{10}\times \frac {7^2}{10^2}\times \sum_{n=0}^{\infty}\left(\frac {7^3}{10^3}\right)^n=$$
$$=\frac 3{10}\times \frac {7^2}{10^2}\times \frac 1{1-\frac {7^3}{10^3}}=\frac {3\times 7^2}{10^3-7^3}=\frac {49}{219}$$
To do it without Geometric Series:
For $i\in \{0,1,2\}$ let $p_i$ denote the probability that you get your first $T$ on a trial with number $\equiv i\pmod 3$.  
Thus, the answer you seek is $p_0$.
Of course, we have $$p_0+p_1+p_2=1$$
Consider $p_1$ first. On the first trial, you either get $T$ (which contributes to $p_1$) or you restart with everything shifted by $1$.  Thus $$p_1=.3+.7\times p_0$$
Similarly, $$p_2=.7\times (.3+.7\times p_0)$$
That gives us three equations in three variables.  We easily solve this system to get $$p_0=\frac {49}{219}\quad p_1=\frac {100}{219}\quad p_2=\frac {70}{219}$$
As I suggested in the comments, it is not a bad idea to do it by Geometric Series if for no reason other than as a check.
A: $S = (0.7)^2\cdot 0.3 + (0.7)^5\cdot 0.3\ + ........+\ (0.7)^{3n-1}\cdot 0.3$
$S = \frac{a_1}{1 - r} = \frac{0.7^2\cdot 0.3}{1 - .343} = \frac{147}{657} = \frac{49}{219}$
$a + b = 268$
