Probability of selecting same factor. Willie Pikette randomly selects a factor of $144$. Betty Wheel selects a factor of $88$. What is the probability that they selected the same number?
This is my incorrect approach (and please feel free to bash at me):
$144$ has $15$ factors in total whereas $88$ has 
$8$ factors.
Because $144 = {2^4}\times{3^2}$ and $88 = {2^3}\times{11}$,
the common factors are related to $2: 1,2,{2^2},{2^3}$ for a total of $4$ factors.
So the probability of choosing the common factor from $144$ is $\frac{4}{15}$
and the probability of choosing a common factor from $88$ is $\frac{4}{8}$.
Using the rules of multiplication, $\frac{4}{15}\times\frac{4}{8} = \frac{2}{15}$ which is approximately $13.3\%$
This isn't the answer; rather it is $3.3\%$
I would gladly appreciate that you guys could not only provide the appropriate analysis and solution, but also point out the error to my solution (hopefully in layman's terms :)).

You've got the factoring part right, but the combinatorical part wrong:


*

*The number of ways to pick a pair of factors is $15\cdot8$

*The number of ways to pick a pair of identical factors is $4$

*Hence the probability of picking a pair of identical factors is $\frac{4}{15\cdot8}$



You have answered correctly for the probability of picking a pair of common factors.
The question, however, is about the probability of picking a pair of identical factors.
