My Courses CRN 40938 Chapter 4 HW - Discrete Random Variable +A! Category: Homework \begin{tabular}{llllllllllllllllllllllllll} \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \end{tabular} Current learning objective: Notation for the Binomial: $$ \mathrm{B}= $$ Binomial Probability Distribution Function Question 20 Practice similar questions Score: 0 of 5 points A baseball player has a batting average of 0.18 . What is the probability that he has exactly 5 hits in his next 7 at bats? (round your answer to four decimal places) The probability is Enter your answer Submit answer Attempts: 0/10 Find this question difficult? Do you know you can practice another version of this question?

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To find the probability that the baseball player has exactly 5 hits in his next 7 at bats, we can use the binomial probability distribution function.

Given:
- Batting average = 0.18
- Number of at bats = 7
- Number of hits = 5

The binomial probability distribution function is given by:
$$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$
where:
- $$ n $$ is the number of trials (at bats in this case)
- $$ k $$ is the number of successes (hits in this case)
- $$ p $$ is the probability of success (batting average in this case)
- $$ \binom{n}{k} $$ is the binomial coefficient, which represents the number of ways to choose $$ k $$ successes out of $$ n $$ trials

Substitute the given values into the binomial probability distribution function:
$$ P(X = 5) = \binom{7}{5} (0.18)^5 (1-0.18)^{7-5} $$

Now, we can calculate the probability using the binomial probability distribution function.
Calculate the value by following steps:
- step0: Calculate:
  $$ { }_{7}C_{5}\times 0.18^{5}\left(1-0.18\right)^{7-5}$$
- step1: Subtract the numbers:
  $$ { }_{7}C_{5}\times 0.18^{5}\times 0.82^{7-5}$$
- step2: Subtract the numbers:
  $$ { }_{7}C_{5}\times 0.18^{5}\times 0.82^{2}$$
- step3: Expand the expression:
  $$\frac{7!}{5!\times \left(7-5\right)!}\times 0.18^{5}\times 0.82^{2}$$
- step4: Subtract the numbers:
  $$\frac{7!}{5!\times 2!}\times 0.18^{5}\times 0.82^{2}$$
- step5: Reduce the fraction:
  $$\frac{7\times 6}{2!}\times 0.18^{5}\times 0.82^{2}$$
- step6: Multiply the numbers:
  $$\frac{42}{2!}\times 0.18^{5}\times 0.82^{2}$$
- step7: Divide the terms:
  $$21\times 0.18^{5}\times 0.82^{2}$$
- step8: Convert the expressions:
  $$21\left(\frac{9}{50}\right)^{5}\times 0.82^{2}$$
- step9: Convert the expressions:
  $$21\left(\frac{9}{50}\right)^{5}\left(\frac{41}{50}\right)^{2}$$
- step10: Simplify:
  $$21\times \frac{9^{5}}{50^{5}}\times \left(\frac{41}{50}\right)^{2}$$
- step11: Multiply the terms:
  $$\frac{1240029}{50^{5}}\times \left(\frac{41}{50}\right)^{2}$$
- step12: Evaluate the power:
  $$\frac{1240029}{50^{5}}\times \frac{41^{2}}{50^{2}}$$
- step13: Multiply the fractions:
  $$\frac{1240029\times 41^{2}}{50^{5}\times 50^{2}}$$
- step14: Multiply:
  $$\frac{2084488749}{50^{7}}$$
The probability that the baseball player has exactly 5 hits in his next 7 at bats is approximately 0.0027 (rounded to four decimal places).
The probability of the baseball player having exactly 5 hits in 7 at bats is approximately 0.0027.

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