Question

a baseball player has a batting average of 0.155. what is the probability that he has exactly 5 hits in his next 7 at bats?
the probability is
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question 19 0/1 pt 3 99 details

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. Here, $n = 7$, $k = 5$, $p = 0.155$.

Step2: Calculate the combination

First, calculate $\binom{7}{5}$. The combination formula is $\binom{n}{k} = \frac{n!}{k!(n - k)!}$. So $\binom{7}{5} = \frac{7!}{5!(7 - 5)!} = \frac{7!}{5!2!} = \frac{7\times6}{2\times1} = 21$.

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

Calculate $p^k = (0.155)^5$ and $(1 - p)^{n - k} = (1 - 0.155)^{7 - 5} = (0.845)^2$.

Step4: Multiply the terms

Now, multiply these three parts together: $P(X = 5) = 21\times(0.155)^5\times(0.845)^2$.

First, calculate $(0.155)^5 \approx 0.155\times0.155\times0.155\times0.155\times0.155 \approx 0.000089$.

Then, calculate $(0.845)^2 = 0.845\times0.845 \approx 0.714025$.

Now, multiply $21\times0.000089 \approx 0.001869$.

Then, multiply $0.001869\times0.714025 \approx 0.001335$.

Answer:

The probability is approximately $0.0013$ (or more precisely, after more accurate calculation:

First, $(0.155)^5 = 0.155^5 \approx 0.00008944$.

$(0.845)^2 = 0.714025$.

$\binom{7}{5} = 21$.

So $21\times0.00008944\times0.714025 \approx 21\times0.0000638 \approx 0.00134$). So the probability is approximately $0.0013$ (or $1.3\times10^{-3}$).