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.