This problem involves calculating the probability that a baseball player with a batting average of 0.305 will get at least 2 hits in 7 at-bats. Since it involves a binomial probability calculation, let's break it down.

Step-by-step approach:

1. Define the binomial distribution parameters:

   • $p = 0.305$ (the probability of getting a hit in one at-bat)
   • $n = 7$ (number of at-bats)
   • We need to find the probability of getting at least 2 hits (so $P(X \geq 2)$).

2. Complement rule: To make the calculation easier, we can use the complement rule: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$ We'll first calculate $P(X = 0)$ and $P(X = 1)$ using the binomial probability formula: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

3. Calculate $P(X = 0)$: $P(X = 0) = \binom{7}{0} (0.305)^0 (0.695)^7 = 1 \times 1 \times (0.695)^7 = 0.695^7 \approx 0.0828$

4. Calculate $P(X = 1)$: $P(X = 1) = \binom{7}{1} (0.305)^1 (0.695)^6 = 7 \times 0.305 \times (0.695)^6$ First, calculate $(0.695)^6 \approx 0.2463$, so: $P(X = 1) \approx 7 \times 0.305 \times 0.2463 \approx 0.5257$

5. Calculate $P(X \geq 2)$: Now, subtract the two probabilities from 1: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.0828 - 0.5257 = 1 - 0.6085 = 0.3915$

Thus, the probability that the player will get at least 2 hits in the game is approximately 0.3915.

Would you like more details on any of these steps?

Here are some related questions you could explore:

1. How does the binomial distribution apply to real-life situations like sports?
2. What is the significance of using the complement rule in probability?
3. How would the probability change if the player's batting average increased to 0.400?
4. What if the player only had 5 at-bats instead of 7? How would that affect the probability?
5. Can we use a different probability distribution for this kind of problem?

Tip: Always double-check calculations involving exponentiation and factorials in binomial distributions to avoid errors.