If a baseball player has a batting average of 0.365​, what is the probability that the player will get the following number of hits in the next four times at​ bat?
​(A) Exactly 2 hits
​(B) At least 2 hits
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Part 1
​(A)​ P(exactly 2 ​hits)almost equals

enter your response here ​(Round to three decimal places as​ needed.)

To solve this problem, we can use the binomial probability formula. The binomial probability formula is given by:

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

where:
- \( n \) is the number of trials (in this case, 4 times at bat),
- \( k \) is the number of successful outcomes (the number of hits),
- \( p \) is the probability of a successful outcome (the batting average, 0.365),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

### Part A: Exactly 2 hits

Let's calculate the probability of getting exactly 2 hits in 4 times at bat.

1. Calculate the binomial coefficient for \( k = 2 \) and \( n = 4 \):
   \[
   \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{24}{4} = 6
   \]

2. Use the binomial probability formula:
   \[
   P(X = 2) = \binom{4}{2} (0.365)^2 (1 - 0.365)^{4-2}
   \]
   \[
   P(X = 2) = 6 \cdot (0.365)^2 \cdot (0.635)^2
   \]

3. Calculate the values:
   \[
   (0.365)^2 = 0.133225
   \]
   \[
   (0.635)^2 = 0.403225
   \]
   \[
   P(X = 2) = 6 \cdot 0.133225 \cdot 0.403225
   \]

4. Multiply the values:
   \[
   P(X = 2) = 6 \cdot 0.05371743125 \approx 0.322
   \]

So, the probability of getting exactly 2 hits is approximately \( 0.322 \) (rounded to three decimal places).

### Part B: At least 2 hits

To find the probability of getting at least 2 hits, we need to sum the probabilities of getting 2, 3, and 4 hits:

\[
P(\text{at least 2 hits}) = P(X = 2) + P(X = 3) + P(X = 4)
\]

1. We've already calculated \( P(X = 2) \approx 0.322 \).

2. Calculate \( P(X = 3) \):
   \[
   P(X = 3) = \binom{4}{3} (0.365)^3 (1 - 0.365)^{4-3}
   \]
   \[
   \binom{4}{3} = 4
   \]
   \[
   P(X = 3) = 4 \cdot (0.365)^3 \cdot (0.635)
   \]
   \[
   (0.365)^3 = 0.048615625
   \]
   \[
   P(X = 3) = 4 \cdot 0.048615625 \cdot 0.635 \approx 0.1238
   \]

3. Calculate \( P(X = 4) \):
   \[
   P(X = 4) = \binom{4}{4} (0.365)^4 (1 - 0.365)^{4-4}
   \]
   \[
   \binom{4}{4} = 1
   \]
   \[
   P(X = 4) = 1 \cdot (0.365)^4 \cdot 1
   \]
   \[
   (0.365)^4 = 0.0177415625
   \]
   \[
   P(X = 4) = 0.0177
   \]

4. Sum the probabilities:
   \[
   P(\text{at least 2 hits}) = 0.322 + 0.124 + 0.018 \approx 0.464
   \]

So, the probability of getting at least 2 hits is approximately \( 0.464 \) (rounded to three decimal places).

**Summary:**
- (A) The probability of getting exactly 2 hits is approximately \( 0.322 \).
- (B) The probability of getting at least 2 hits is approximately \( 0.464 \).

Do you want more details or have any questions?

### Related Questions:

1. What is the probability of getting exactly 3 hits?
2. What is the probability of getting exactly 1 hit?
3. What is the probability of getting no hits?
4. What is the probability of getting at most 2 hits?
5. How does the probability change if the batting average increases to 0.400?
6. How does the probability change if the number of at bats increases to 5?
7. What is the cumulative probability distribution for this scenario?
8. How can the binomial probability formula be applied to other sports statistics?

**Tip:** When dealing with binomial probabilities, remember that the sum of probabilities for all possible outcomes should equal 1.

Learn how to calculate the probability of a baseball player getting exactly 2 hits and at least 2 hits in the next four times at bat. This problem involves applying the binomial probability formula with a batting average of 0.365.

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