based on past results, a batter knows that the opposing pitcher throws a fastball 75% of the time and a…

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.\nsample problem\np(4 fastballs and 4 curveballs)\n=8c4(3/4)^4(1/4)^4\n≈70(0.0012359)\n≈0.087\n≈8.7%\np(7 fastballs and 1 curveball)\n> enter the answer in the space provided. use numbers instead of words.\n%
Answer
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%