An Olympic archer is able to hit the bull's-eye 80% of the time. Assume each shot is independent of the others. If she shoots 10 arrows, what's the probability that she hits the bull's-eye more often then she misses?
[Math] Probability in Statistics
probability
Related Solutions
Let $q_i = P(X=i), i=0,1,2,\ldots$. Assume $i\gt 0$ and consider the outcome of the first dart. To obtain $i$ bullseyes it must either be a bullseye, in which case we need a further $i-1$ bullseyes, or a non-bullseye hit, in which case we still need $i$ bullseyes. Therefore, using the law of total probability,
\begin{eqnarray*} q_i &=& \dfrac{p}{2}q_{i-1} + \dfrac{1-p}{2}q_i \\ \therefore (p+1)q_i &=& pq_{i-1} \\ q_i &=& \dfrac{p}{p+1}q_{i-1} \\ q_i &=& \left(\dfrac{p}{p+1}\right)^iq_{0}\qquad\text{solving the simple recurrence relation.} \\ \end{eqnarray*}
We now require $q_0$. To obtain $0$ bullseyes we need $0$ or more non-bullseye hits followed by a miss. Therefore,
\begin{eqnarray*} q_0 &=& \dfrac{1}{2} \sum_{n=0}^\infty{\left(\dfrac{1-p}{2}\right)^n} \\ &=& \dfrac{1}{2}\cdot \dfrac{1}{1-\frac{1-p}{2}} \qquad\text{using geometric series formula} \\ &=& \dfrac{1}{p+1}. \end{eqnarray*}
Therefore, $$q_i = \dfrac{p^i}{\left(p+1\right)^{i+1}}, \quad i=0,1,2,\ldots.$$
You and Mosteller are calculating slightly different quantities.
Let's look at what you are calculating for case 1:
You write $P(\text{A survives in case A shots B})$ which means $P\text{(A survives given A kills B}) \equiv P(\text{A survives|A kills B})$
But in the right hand side of your equation you are calculating $P(\text{A survives}\ \cap\ \text{A kills B})$ which is equal to $P(\text{A survives|A kills B})\cdot P(\text{A kills B}) = \frac{3}{13}\cdot 0.3$
So you are calculating the intersection of two events (event 1 and event 2) instead of the conditional (event 1 given event 2). Note that if you use the right name/description for the probability you are calculating then your calculations are agreeing with Mosteller.
The same goes for case 2. You are calculating $P(\text{A survives}\ \cap \ \text{A shoots at B and misses})$
Now that we have named the probabilities more accurately, you can ask yourself: How do they help me in answering the question? Isn't it more useful if I know the conditional probabilities instead? Yes it is. This is what will help you decide on the optimal action for A.
So as Monteller says: If A has to pick a target, he has to go for B. If A misses B, then B kills C on the next round, and on the round after that A has a single chance at B. So A survives with probability $0.3$. If A kills B then we find that the probability of surviving a duel with C (where C goes first) is $\frac{3}{13}$. So we see it's better to miss. And indeed A can choose to deliberately miss, and this is what he should do.
Finally here's another way to find A's survival probability against a duel with C. Let's define two probabilities, $P_A$ and $P_C$: $$ P_A \equiv P(\text{A shoots first and A survives at the end}) \\ P_C \equiv P(\text{C shoots first and C survives at the end}) $$
Now note the relationship between the two, that creates a $2\times 2$ system that can be easily solved: $$ \left. \begin{array}{} P_A = 0.3 + 0.7\cdot (1-P_C)\\ P_C = 0.5 + 0.5\cdot (1-P_A) \end{array} \right\} \iff \begin{array}{} P_A = \frac{6}{13}\\ P_C = \frac{10}{13} \end{array} $$
What we care about in our scenario is person A surviving given that person C goes first, which is equal to $1-P_C = \frac{3}{13}$
Best Answer
If she hits the bullseye more often than she misses, then she would've hit the bullseye more than $5$ times. You can compute each case separately (once for each of $6$, $7$, $8$, $9$, and $10$ times she hits the bullseye) then add them. Since you know how to use the binomial distribution formulas, you can do that. I'll do the first one for you:
Hitting the bullseye $6$ times means missing it $4$ times: $$P(6 \text{ hits})={10\choose 6}\times 0.8^{6}\times0.2^4=0.088$$ Can you continue with the other $4$ cases?