The question is the following:
A baseball player is said to “hit for the cycle” if he has a single, a double, a triple, and a home run all in one game. Suppose these four types of hits have probabilities $1/16, 1/4, 1/5,$ and $1/24$. What is the probability of hitting for the cycle if he gets to bat eight times?
Some people said that my procedure and the answer is wrong.
I tried to do it like the following. Did I go wrong somewhere?
The way I proceeded
Given that,
P(Single)= $\frac{1}{16}$
P(Double)= $\frac{1}{4}$
P(Triple)= $\frac{1}{5}$
P(Home Run) = $\frac{1}{24}$
As there are $4$ types of hits, so the value of $r$ would be $4$.
Here,
$r=4, n=8$
$\therefore$ The probability of hitting for the cycle if he gets to bat for $8$ times is,
\begin{align*}
P(x=8) & = {8\choose4}\hspace{.1cm} \cdot P(\text{Single}) \cdot P(\text{Double}) \cdot P(\text{Triple}) \cdot P(\text{Home run})\\
& =\hspace{.1cm}{8\choose 4}\hspace{.1cm} \cdot \frac{1}{16} \cdot \frac{1}{4} \cdot \frac{1}{5} \cdot \frac{1}{24}\\
& =\frac{70}{7680}\\
& \hspace{.1cm}\equiv 0.91\%
\end{align*}
Best Answer
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.