probability
A baseball player has a 30% chance of hitting at any bat, what is the probability that this player gets hits on each of his next 2 at bats and then makes an out (misses) at the next bat?
Would it be 0.063? Because P(hits) = 3/10, and P(miss) = 7/10, I did this: 0.3 x 0.3 x 0.7. However, I'm not sure if this is the right equation to use?
Other possible answers: 0.670, 0.090, or 0.210
Thank you.
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 $1-P_C = \frac{3}{13}$
The probability that the baseball players hits for the cycle is found by subtracting the probability that he obtains no singles or no doubles or no triples or no home runs from $1$.
Let $S$ be the event that he hits at least one single.
Let $D$ be the event that he hits at least one double.
Let $T$ be the event that he hits at least one triple.
Let $H$ be the event that he hits at least one home run.
The probability that the baseball player hits for the cycle in eight plate appearances is $$\Pr(\text{hits for cycle}) = 1 - \Pr(S' \cup D' \cup T' \cup H')$$ where \begin{align*} & \Pr(S' \cup D' \cup T' \cup H')\\ & \quad = \Pr(S') + \Pr(D') + \Pr(T') + \Pr(H')\\ & \qquad - \Pr(S' \cap D') - \Pr(S' \cap T') - \Pr(S' \cap H') - \Pr(D' \cap T') - \Pr(D' \cap H') + \Pr(T' \cap H')\\ & \quad\qquad + \Pr(S' \cap D' \cap T') + \Pr(S' \cap D' \cap H') + \Pr(S' \cap T' \cap H') + \Pr(D' \cap T' \cap H')\\ & \qquad\qquad - \Pr(S' \cap D' \cap T' \cap H') \end{align*} We are given the probabilities \begin{align*} \Pr(\text{single}) & = \frac{1}{16}\\ \Pr(\text{double}) & = \frac{1}{4}\\ \Pr(\text{triple}) & = \frac{1}{5}\\ \Pr(\text{home run}) &= \frac{1}{24} \end{align*} for a plate appearance.
$\Pr(S')$: Since $\Pr(\text{single}) = \frac{1}{16}$, the probability of not obtaining a single in a plate appearance is
$$\Pr(\text{no single}) = 1 - \frac{1}{16} = \frac{15}{16}$$
Assuming independence, the probability of not obtaining a single in eight plate appearances is
$$\Pr(S') = \left(\frac{15}{16}\right)^8$$
By similar reasoning, \begin{align*} \Pr(D') & = \left(1 - \frac{1}{4}\right)^8 = \left(\frac{3}{4}\right)^8\\ \Pr(T') & = \left(1 - \frac{1}{5}\right)^8 = \left(\frac{4}{5}\right)^8\\ \Pr(H') & = \left(1 - \frac{1}{24}\right)^8 = \left(\frac{23}{24}\right)^8 \end{align*}
Next, we note that for a plate appearance, the events single, double, triple, and home run are mutually exclusive.
$\Pr(S' \cap D')$: Since the events single and double are mutually exclusive, the probability of obtaining a single or a double in a plate appearance is
$$\Pr(\text{single} \cup \text{double}) = \Pr(\text{single}) + \Pr(\text{double}) = \frac{1}{16} + \frac{1}{4} = \frac{5}{16}$$
Hence, the probability of obtaining neither a single nor a double in a plate appearance is
$$\Pr(\text{neither a single nor a double}) = 1 - \frac{5}{16} = \frac{11}{16}$$
Thus, the probability of obtaining neither a single nor a double in eight plate appearances is
$$\Pr(S' \cap D') = \left(\frac{11}{16}\right)^8$$
By similar reasoning,
\begin{align*}
\Pr(S' \cap T') & = \left(1 - \frac{1}{16} - \frac{1}{5}\right)^8 = \left(1 - \frac{21}{80}\right)^8 = \left(\frac{59}{80}\right)^8\\
\Pr(S' \cap H') & = \left(1 - \frac{1}{16} - \frac{1}{24}\right)^8 = \left(1 - \frac{5}{48}\right)^8 = \left(\frac{43}{48}\right)^8\\
\Pr(D' \cap T') & = \left(1 - \frac{1}{4} - \frac{1}{5}\right)^8 = \left(1 - \frac{9}{20}\right)^8 = \left(\frac{11}{20}\right)^8\\
\Pr(D' \cap H') & = \left(1 - \frac{1}{4} - \frac{1}{24}\right)^8 = \left(1 - \frac{7}{24}\right)^8 = \left(\frac{17}{24}\right)^8\\
\Pr(T' \cap H') & = \left(1 - \frac{1}{5} - \frac{1}{24}\right)^8 = \left(1 - \frac{29}{120}\right)^8 = \left(\frac{91}{120}\right)^8
\end{align*}
$\Pr(S' \cap D' \cap T')$: Since the events single, double, and triple are mutually exclusive, the probability of obtaining a single, double, or triple in a plate appearance is $$\Pr(\text{single} \cup \text{double} \cup \text{triple}) = \Pr(\text{single}) + \Pr(\text{double}) + \Pr(\text{triple}) = \frac{1}{16} + \frac{1}{4} + \frac{1}{5} = \frac{41}{80}$$ Thus, the probability of obtaining neither a single nor a double nor a triple in a plate appearance is $$\Pr(\text{neither a single nor a double nor a triple}) = 1 - \frac{41}{80} = \frac{39}{80}$$ Hence, the probability of obtaining no singles, doubles, or triples in eight plate appearances is $$\Pr(S' \cap D' \cap T') = \left(\frac{39}{80}\right)^8$$ By similar reasoning, \begin{align*} \Pr(S' \cap D' \cap H') & = \left(1 - \frac{1}{16} - \frac{1}{4} - \frac{1}{24}\right) = \left(1 - \frac{17}{48}\right)^8 = \left(\frac{31}{48}\right)^8\\ \Pr(S' \cap T' \cap H') & = \left(1 - \frac{1}{16} - \frac{1}{5} - \frac{1}{24}\right)^8 = \left(1 - \frac{73}{240}\right)^8 = \left(\frac{167}{240}\right)^8\\ \Pr(D' \cap T' \cap H') & = \left(1 - \frac{1}{4} - \frac{1}{5} - \frac{1}{24}\right)^8 = \left(1 - \frac{59}{120}\right)^8 = \left(\frac{61}{120}\right)^8 \end{align*}
$\Pr(S' \cap D' \cap T' \cap H')$: Since the four events are mutually exclusive, the probability of obtaining a hit in a single plate appearance is \begin{align*} \Pr(\text{single} \cup \text{double} \cup \text{triple} \cup \text{home run}) & = \Pr(\text{single}) + \Pr(\text{double} + \Pr(\text{triple}) + \Pr(\text{home run})\\ & = \frac{1}{16} + \frac{1}{4} + \frac{1}{5} + \frac{1}{24}\\ & = \frac{133}{240} \end{align*} Hence, the probability of not getting a hit in a plate appearance is $$\Pr(\text{neither a single nor a double nor a triple nor a home run}) = 1 - \frac{133}{240} = \frac{107}{240}$$ Hence, the probability that a player does not get a hit in eight plate appearances is $$\Pr(S' \cap D' \cap T' \cap H') = \left(\frac{107}{240}\right)^8$$ Therefore, the probability that a player hits for the cycle in eight plate appearances is \begin{align*} \Pr(\text{hits for cycle}) & = 1 - \left(\frac{15}{16}\right)^8 - \left(\frac{3}{4}\right)^8 - \left(\frac{4}{5}\right)^8 - \left(\frac{23}{24}\right)^8\\ & \quad + \left(\frac{11}{16}\right)^8 + \left(\frac{59}{80}\right)^8 + \left(\frac{43}{48}\right)^8 + \left(\frac{11}{20}\right)^8 + \left(\frac{17}{24}\right)^8 + \left(\frac{91}{120}\right)^8\\ & \qquad - \left(\frac{39}{80}\right)^8 - \left(\frac{31}{48}\right)^8 - \left(\frac{167}{240}\right)^8 - \left(\frac{61}{120}\right)^8\\ & \quad\qquad + \left(\frac{107}{240}\right)^8 \end{align*}
The numbers used in this problem are rather unrealistic. Triples are rare events. The most triples hit in a season in Major League Baseball is $36$ by Owen Wilson in the year 1912 C.E. and it took him $643$ plate appearances to do it. Since hitting a triple is a rare event, hitting for the cycle is a still rarer event. In fact, no Major League Baseball player has hit for the cycle more than three times in his entire career.
Best Answer
Yes, your answer is correct.
The probability of a hit is $30\%=0.3$, so the probability of a miss is $1-30\%=70\%=0.7$.
Assuming that all bats are independent (the outcome of one bat does not affect the other bats), then the probability of hitting the next two bats and missing the third is $0.3\times0.3\times0.7=0.063=6.3\%$.
Note that it is important that the question specified that the player hits his next 2 bats and misses the next bat. If the question did not specify the order and you were just finding the probability of two misses out of three bats, the answer is different.