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)
=8c4(3/4)^4(1/4)^4
≈70(0.0012359)
≈0.087
≈8.7%
p(7 fastballs and 1 curveball)

enter the answer in the space provided. use numbers instead of words.

%

Explanation:

Step1: Calculate the binomial coefficient

We use the binomial - probability formula \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successful trials, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n = 8\), the probability of a fast - ball \(p = 0.75\) and the probability of a curve - ball \(1 - p=0.25\). For the case of 7 fast - balls and 1 curve - ball, \(k = 7\). First, calculate \(C(8,7)=\frac{8!}{7!(8 - 7)!}=\frac{8!}{7!1!}=\frac{8\times7!}{7!×1}=8\).

Step2: Calculate the probability

Then, \(P(7\text{ fast - balls and }1\text{ curve - ball})=C(8,7)\times(0.75)^{7}\times(0.25)^{1}\).
\((0.75)^{7}=0.133483764\), \((0.25)^{1}=0.25\), and \(C(8,7) = 8\).
So \(P(7\text{ fast - balls and }1\text{ curve - ball})=8\times0.133483764\times0.25\).
\(P(7\text{ fast - balls and }1\text{ curve - ball})=8\times0.133483764\times0.25 = 0.266967528\approx26.7\%\).

Answer:

26.7%