Question

a high school baseball player has a 0.202 batting average. in one game, he gets 8 at bats. what is the probability he will get at least 3 hits in the game?

Explanation:

Step1: Identify Distribution

This is a binomial probability problem. Let \( n = 8 \) (number of trials), \( p = 0.202 \) (probability of success), and we want \( P(X \geq 3) \), where \( X \) is the number of hits. We use the complement rule: \( P(X \geq 3)=1 - P(X = 0)-P(X = 1)-P(X = 2) \). The binomial probability formula is \( P(X = k)=\binom{n}{k}p^{k}(1 - p)^{n - k} \), where \( \binom{n}{k}=\frac{n!}{k!(n - k)!} \).

Step2: Calculate \( P(X = 0) \)

\( \binom{8}{0}=\frac{8!}{0!8!}=1 \), \( p^{0}=1 \), \( (1 - p)^{8}=(1 - 0.202)^{8}=(0.798)^{8}\approx0.163 \). So \( P(X = 0)=1\times1\times0.163 = 0.163 \).

Step3: Calculate \( P(X = 1) \)

\( \binom{8}{1}=\frac{8!}{1!7!}=8 \), \( p^{1}=0.202 \), \( (1 - p)^{7}=(0.798)^{7}\approx0.204 \). So \( P(X = 1)=8\times0.202\times0.204\approx8\times0.0412 = 0.3296 \).

Step4: Calculate \( P(X = 2) \)

\( \binom{8}{2}=\frac{8!}{2!6!}=\frac{8\times7}{2\times1}=28 \), \( p^{2}=(0.202)^{2}=0.0408 \), \( (1 - p)^{6}=(0.798)^{6}\approx0.256 \). So \( P(X = 2)=28\times0.0408\times0.256\approx28\times0.0104 = 0.2912 \).

Step5: Calculate \( P(X \geq 3) \)

\( P(X \geq 3)=1-(0.163 + 0.3296 + 0.2912)=1 - 0.7838 = 0.2162 \) (approximate, more precise calculation of powers may vary slightly).

Answer:

\(\approx0.216\) (or more precise value depending on calculator precision, e.g., using more accurate \( (0.798)^{k} \) values: \( (0.798)^8\approx0.1627 \), \( (0.798)^7\approx0.2039 \), \( (0.798)^6\approx0.2555 \), then \( P(X=0)\approx0.1627 \), \( P(X = 1)\approx8\times0.202\times0.2039\approx0.328 \), \( P(X = 2)\approx28\times0.0408\times0.2555\approx28\times0.0104 = 0.291 \), so \( 1-(0.1627 + 0.328+0.291)=1 - 0.7817 = 0.2183\approx0.218 \))