Question

based on past results, a batter knows that the opposing pitcher throws a fastball 75% of the time and a curveball 25% of the time. suppose the batter sees 8 pitches during a particular at - bat. determine each probability. round your answers to the nearest tenth of a percent if necessary. sample problem p(4 fastballs and 4 curveballs) = _8c_4(\frac{3}{4})^4(\frac{1}{4})^4≈70(0.0012359)≈0.087≈8.7% p(5 curveballs and 3 fastballs) enter the answer in the space provided. use numbers instead of words.

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, the probability of a fast - ball $p = 0.75$ and the probability of a curve - ball $1 - p=0.25$, and the number of pitches $n = 8$.

Step2: Calculate $C(n,k)$ for $n = 8$ and $k = 3$

$C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8!}{3!5!}=\frac{8\times7\times6}{3\times2\times1}=56$.

Step3: Calculate $p^{k}$ and $(1 - p)^{n - k}$

$p^{k}=(0.75)^{3}$ and $(1 - p)^{n - k}=(0.25)^{5}$.
$(0.75)^{3}=0.75\times0.75\times0.75 = 0.421875$ and $(0.25)^{5}=0.25\times0.25\times0.25\times0.25\times0.25=0.0009765625$.

Step4: Calculate the probability

$P(X = 3)=C(8,3)\times(0.75)^{3}\times(0.25)^{5}$
$P(X = 3)=56\times0.421875\times0.0009765625$
$P(X = 3)=56\times0.0004119873047$
$P(X = 3)\approx0.023$
To convert to a percentage, multiply by 100: $0.023\times100 = 2.3\%$

Answer:

$2.3$