edit: solved
Hi chickeneaterguy,
(a)
The probability Jennifer receives no offer is
$\displaystyle \left(\frac{2}{3}\right)^4$
Hence subtract that from 1 for the probability she gets at least one offer,
or sum all the seperate ways she can get 1, 2, 3 or 4 offers.
(b)
$\displaystyle 1-\left(\frac{2}{3}\right)^n\ >\ 0.9$
$\displaystyle 1-0.9\ >\ \left(\frac{2}{3}\right)^n$
$\displaystyle log(0.1)\ >\ log\left(\frac{2}{3}\right)^n$
$\displaystyle log(0.1)\ >\ nlog\left(\frac{2}{3}\right)$
As log(2/3) is negative, reverse the inequality to solve for n
$\displaystyle \frac{log(0.1)}{log\left(\frac{2}{3}\right)}\ <\ n$
$\displaystyle n\ >\ 5.67\ \Rightarrow\ n\ \ge\ 6$
Sure,
when the value we are looking for is in the index position,
we can extract using logs, as
$\displaystyle log\left(x^n\right)=n(logx)$
then we can find n.
You can also try some experimental values in this case,
as we need not expect a large value of n to be the solution.
$\displaystyle 0.1\ >\ \left(\frac{2}{3}\right)^n$
$\displaystyle \frac{2}{3}=0.67$
$\displaystyle \left(\frac{2}{3}\right)^2=0.44$
$\displaystyle \left(\frac{2}{3}\right)^3=0.296$
$\displaystyle \left(\frac{2}{3}\right)^4=0.1975$
$\displaystyle \left(\frac{2}{3}\right)^5=0.1316$
$\displaystyle \left(\frac{2}{3}\right)^6=0.08779$