Mark has a batting average of 0.155. Let X be the number of hits in his next 18 at bats.
What is the distribution of X? X  (,)
Please show the following answers to 4 decimal places.
What is the probability that he has exactly 6 hits? 
What is the probability that he has less than 6 hits? 
What is the probability that he has at most 6 hits? 
What is the probability that he has between 3 and 9 (including 3 and 9) hits? 

We can model this problem as a **binomial distribution**, where:

- \( X \) is the number of hits in 18 at-bats.
- Mark's probability of a hit in any given at-bat is \( p = 0.155 \).
- The number of at-bats \( n = 18 \).

The binomial distribution is denoted as \( X \sim \text{Binomial}(n, p) \), so:

\[
X \sim \text{Binomial}(18, 0.155)
\]

Now let's answer the given questions using this distribution.

### 1. **The probability that Mark has exactly 6 hits**
This is given by the binomial probability mass function (PMF):

\[
P(X = 6) = \binom{18}{6} \cdot (0.155)^6 \cdot (1 - 0.155)^{18-6}
\]

### 2. **The probability that he has less than 6 hits**
This means calculating the cumulative probability for \( X = 0, 1, 2, 3, 4, 5 \):

\[
P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\]

### 3. **The probability that he has at most 6 hits**
This is the cumulative probability for \( X = 0, 1, 2, 3, 4, 5, 6 \):

\[
P(X \leq 6) = P(X < 6) + P(X = 6)
\]

### 4. **The probability that he has between 3 and 9 hits (including 3 and 9)**
This is the cumulative probability for \( X = 3, 4, 5, 6, 7, 8, 9 \):

\[
P(3 \leq X \leq 9) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
\]

Let's calculate these probabilities.Here are the calculated probabilities, rounded to four decimal places:

1. **The probability that Mark has exactly 6 hits**: \( P(X = 6) = 0.0341 \)
2. **The probability that he has less than 6 hits**: \( P(X < 6) = 0.9518 \)
3. **The probability that he has at most 6 hits**: \( P(X \leq 6) = 0.9859 \)
4. **The probability that he has between 3 and 9 hits (inclusive)**: \( P(3 \leq X \leq 9) = 0.5440 \)

Would you like further details or have any additional questions?

Here are some related questions for further exploration:

1. What is the probability that Mark gets no hits in 18 at-bats?
2. What is the expected number of hits in 18 at-bats?
3. What is the standard deviation for the number of hits?
4. How does increasing the number of at-bats affect the probability of getting exactly 6 hits?
5. What is the probability of getting at least 1 hit in the next 18 at-bats?

**Tip:** For small probabilities and large trials, approximating the binomial distribution with the Poisson distribution is often useful.

Mark has a batting average of 0.155. Let X be the number of hits in his next 18 at bats. What is the distribution of X? What is the probability that he has exactly 6 hits? What is the probability that he has less than 6 hits? What is the probability that he has at most 6 hits? What is the probability that he has between 3 and 9 (including 3 and 9) hits?

Explore the probability of Mark getting hits in his next 18 at-bats using binomial distribution. This detailed guide shows how to calculate the probability of exactly 6 hits, less than 6 hits, at most 6 hits, and hits between 3 and 9 (inclusive). Suitable for high school students studying probability and statistics.

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