Question

a tennis player makes a successful first serve 51% of the time. if she serves 9 times, what is the probability that she gets exactly 3 successful first serves in? assume that each serve is independent of the others.

a. 0.154
b. 0.133
c. 0.0635
d. 0.00184

Explanation:

Step1: Identify binomial 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 $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine values of $n$, $k$, and $p$

Here, $n = 9$ (number of serves), $k = 3$ (number of successful serves), and $p=0.51$ (probability of a successful first - serve), and $1 - p = 1-0.51 = 0.49$.

Step3: Calculate the combination $C(n,k)$

$C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9!}{3!6!}=\frac{9\times8\times7}{3\times2\times1}=84$.

Step4: Calculate the probability

$P(X = 3)=C(9,3)\times(0.51)^{3}\times(0.49)^{6}$
$P(X = 3)=84\times(0.51)^{3}\times(0.49)^{6}$
$P(X = 3)=84\times0.132651\times0.01680664$
$P(X = 3)\approx0.184$.

Answer:

None of the provided options are correct. The calculated probability is approximately $0.184$.