The 2009 batting champion in a national baseball league has a batting average of 0.349. This can be interpreted as a probability of 0.349 of getting a hit whenever he bats. Assume that each time at bat is an independent event. Suppose he goes to bat 4 times in a game.
​(a) Find the probability distribution for the number of hits.
​(b) What is the expected number of hits that the champion gets in a​ game?
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Part 1
​(a) Select the correct choice below and fill in the answer boxes within your choice.
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To solve this problem, we can model it using a **binomial distribution**. The number of hits in a game follows a binomial distribution because:

1. The player either gets a hit or not, which is a Bernoulli trial.
2. The probability of success (getting a hit) is constant for each trial, \( p = 0.349 \).
3. The number of at-bats (trials) is fixed at \( n = 4 \).

### Part (a) - Find the Probability Distribution

The probability mass function for a binomial distribution is given by:

\[
P(x) = \binom{n}{x} p^x (1 - p)^{n - x}
\]

Where:
- \( n = 4 \) is the number of trials (at-bats),
- \( x \) is the number of successes (hits),
- \( p = 0.349 \) is the probability of success (getting a hit),
- \( \binom{n}{x} \) is the binomial coefficient, calculated as \( \binom{n}{x} = \frac{n!}{x!(n-x)!} \).

Now, we calculate the probabilities for each possible value of \( x \) (number of hits) from 0 to 4.

#### 1. \( P(x = 0) \)

\[
P(0) = \binom{4}{0} (0.349)^0 (1 - 0.349)^4 = 1 \times 1 \times (0.651)^4 = 0.1796
\]

#### 2. \( P(x = 1) \)

\[
P(1) = \binom{4}{1} (0.349)^1 (1 - 0.349)^3 = 4 \times 0.349 \times (0.651)^3 = 0.3841
\]

#### 3. \( P(x = 2) \)

\[
P(2) = \binom{4}{2} (0.349)^2 (1 - 0.349)^2 = 6 \times (0.349)^2 \times (0.651)^2 = 0.3081
\]

#### 4. \( P(x = 3) \)

\[
P(3) = \binom{4}{3} (0.349)^3 (1 - 0.349)^1 = 4 \times (0.349)^3 \times 0.651 = 0.1112
\]

#### 5. \( P(x = 4) \)

\[
P(4) = \binom{4}{4} (0.349)^4 (1 - 0.349)^0 = 1 \times (0.349)^4 \times 1 = 0.0169
\]

Thus, the probability distribution for the number of hits is:

| x   | 0     | 1     | 2     | 3     | 4     |
|-----|-------|-------|-------|-------|-------|
| P(x)| 0.1796| 0.3841| 0.3081| 0.1112| 0.0169|

### Part (b) - Expected Number of Hits

The expected value \( E(X) \) for a binomial distribution is calculated as:

\[
E(X) = n \cdot p
\]

Where \( n = 4 \) and \( p = 0.349 \):

\[
E(X) = 4 \cdot 0.349 = 1.396
\]

Thus, the expected number of hits that the player gets in a game is **1.396**.

Would you like any further details or clarification?

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Here are 5 related questions to explore:

1. How would the probabilities change if the player went to bat 5 times instead of 4?
2. What is the probability of the player getting exactly 2 hits in a game?
3. How do you calculate the variance for a binomial distribution?
4. What if the player's batting average improved to 0.400? How would that affect the expected number of hits?
5. Can you interpret the meaning of the expected value in the context of this problem?

**Tip**: The sum of all probabilities in a probability distribution must always equal 1. This can help in verifying your calculations.

The 2009 batting champion in a national baseball league has a batting average of 0.349. This can be interpreted as a probability of 0.349 of getting a hit whenever he bats. Assume that each time at bat is an independent event. Suppose he goes to bat 4 times in a game. (a) Find the probability distribution for the number of hits. (b) What is the expected number of hits that the champion gets in a game?

Learn how to find the probability distribution of a baseball player's hits when their batting average is 0.349, assuming 4 at-bats in a game. The solution covers binomial distribution calculations, including the probability mass function and the expected number of hits. Ideal for Grades 11-12 students studying probability.

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