Find probability that Rahul examines fewer policies than toby 
$$ \underline{ \bf Attempt} $$
Let $X$ and $Y$ be the policies examined by rahul and toby, respectively. By the manner this problem is phrased we have two Bernoulli r.v. For instance, $X$ has probability success of $0.1$, thus 
$$ p_X(x) = {n \choose x} 0.1^x 0.8^{n-x}$$
and similarly
$$ p_Y(y) = {n \choose y} 0.2^y 0.9^{n-y} $$
Now, since $X$ and $Y$ are given to be independent, we have 
$$ p_{XY}(x,y) = {n \choose x} {n \choose y}  0.1^x 0.8^{n-x} 0.2^y 0.9^{n-y} $$
Now, we want to find 
$$ P(X<Y) $$
But, for discrete case how do we compute this? Do we compute 
$$ \sum_{x<y} p_{XY}(x,y) $$
Perhaps my solution is not really the right to do this problem?
 A: Disclaimer
Here I assume that the number of policies are infinite.
Raul point of view


*

*Examine the first policy. If it meets the requirement, Raul stops. In this case the probability is $0.1$.  Then the probability that he examines $1$ policy is $0.1$.

*Otherwise, with probability $0.9$, he examines the second. He stops with probability $0.1$. Then the probability that he examines $2$ policies is $0.1 \cdot 0.9$.

*Otherwise, with probability $0.9$, he examines the third. He stops with probability $0.1$. Then the probability that he examines $3$ policies is $0.1 \cdot 0.9^2$.


In general:
$$p_{X}(x) = 0.1 \cdot 0.9^{x-1},$$
is the probability that Rahul wil examine exactly $x$ policies. For Toby, we have a similar result:
$$p_{Y}(y) = 0.2 \cdot 0.8^{y-1}.$$
The calculations
In this case:
$$\begin{eqnarray}p(X < Y) & = & \sum_{y=2}^{+\infty} \sum_{x=1}^{y-1}p_{X}(x)p_{Y}(y) \\
& =& \frac{0.1 \cdot 0.2}{0.9 \cdot 0.8} \sum_{y=2}^{+\infty} \sum_{x=1}^{y-1} 0.9^x 0.8^y \\
& =& \frac{1}{36} \sum_{y=2}^{+\infty} 0.8^y \left(\sum_{x=1}^{y-1} 0.9^x\right).
\end{eqnarray}
$$
Recalling that:
$$\sum_{i=1}^{m-1} r^i = \frac{r- r^m}{1-r},$$
we get:
\begin{eqnarray}p(X < Y) & = & \frac{1}{36} \sum_{y=2}^{+\infty} 0.8^y \frac{0.9 - 0.9^y}{1-0.9} \\
& = & \frac{1}{36 \cdot 0.1} \left(0.9\sum_{y=2}^{+\infty} 0.8^y - \sum_{y=2}^{+\infty} 0.72^y\right). \\
\end{eqnarray}
Furthermore, since:
$$\sum_{i=2}^{+\infty} r^i = \frac{r^2}{1-r} ~\text{for}~ |r|<1,$$
then:
$$\begin{eqnarray}p(X < Y) & = &\frac{1}{36 \cdot 0.1} \left(0.9\frac{0.8^2}{1-0.8} - \frac{0.72^2}{1-0.72}\right) \simeq 0.2857. \\
\end{eqnarray}
$$
Final remarks
IMHO, the problem requires the use of geometrical random variable, not binomial. The binomial is used to establish the probability that in a (finite) group of $n$ policies you will find exactly $x$ (or $y$) good policies. This is not coherent with the request of the problem. The problem has a "stop" condition: you examine until you find a good policy.
On the other hand, to use the geometrical random variable, you need to assume that the number of policies to be examined is infinite. This problem can be solved. Indeed, if $n$ is the number of available policies to Rahul (and to Toby), you get that:
$$p_{X}(x) = \begin{cases}
0.1 \cdot 0.9^{x-1} & \text{if}~ x \leq n-1\\
\displaystyle \frac{0.9^n-0.1}{0.9} & \text{if}~ x = n
\end{cases},$$
and
$$p_{Y}(y) = \begin{cases}
0.2 \cdot 0.8^{x-1} & \text{if}~ x \leq n-1\\
\displaystyle \frac{0.8^n-0.2}{0.8} & \text{if}~ x = n
\end{cases}.$$
The terms for $x=n$ are used to say that, when you arrive to $n$-th policy, whether it is good or not, you will stop, since you don't have other policies to examine.
Thereafter, using these probabilities, you can act in a similar way to obtain the probability $P(X < Y)$.
