Question

the best player on a baseball team has a 0.468 batting average. if all other factors are ignored, what is the probability that if the player comes to bat eleven times he gets exactly eight hits in todays game?

the probability that the player gets exactly eight hits is
(round to the four decimal places as needed.)

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. The binomial probability formula is $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $\binom{n}{k} = \frac{n!}{k!(n - k)!}$.
Here, $n = 11$ (number of at - bats), $k = 8$ (number of hits), and $p = 0.468$ (batting average, probability of getting a hit in a single at - bat).

Step2: Calculate the combination $\binom{n}{k}$

First, calculate $\binom{11}{8}$. Since $\binom{n}{k}=\binom{n}{n - k}$, $\binom{11}{8}=\binom{11}{3}$.
$\binom{11}{3}=\frac{11!}{3!(11 - 3)!}=\frac{11!}{3!8!}=\frac{11\times10\times9}{3\times2\times1}=165$

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

$p^k=(0.468)^8$
$(1 - p) = 1 - 0.468 = 0.532$
$(1 - p)^{n - k}=(0.532)^{11 - 8}=(0.532)^3$

Step4: Calculate the probability

$P(X = 8)=\binom{11}{8}(0.468)^8(0.532)^3$
First, calculate $(0.468)^8\approx0.468\times0.468\times0.468\times0.468\times0.468\times0.468\times0.468\times0.468\approx0.00123$
$(0.532)^3 = 0.532\times0.532\times0.532\approx0.149$
Then, $P(X = 8)=165\times0.00123\times0.149$
$165\times0.00123 = 0.20295$
$0.20295\times0.149\approx0.0302$

Answer:

0.0302