Let's solve this step by step.

Exercise 1:

Part 1: Justification for Binomial Distribution

The random variable $C$, which counts the number of successful hits during the training session, follows a binomial distribution because:

1. Each shot has only two possible outcomes: hitting the target (success) or missing (failure).
2. The probability of success (hitting the target) is constant for each shot and is given as $p = \frac{3}{4}$.
3. The shots are independent, meaning the outcome of one shot does not influence the outcomes of others.
4. The total number of trials is fixed, $n = 60$.

Thus, $C \sim \mathcal{B}(n, p)$, where:

• $n = 60$ (number of trials),
• $p = \frac{3}{4} = 0.75$ (probability of success).

Part 2: Probability that $C \geq 50$

We want to calculate $P(C \geq 50)$. Using the binomial distribution formula or approximations: $P(C \geq 50) = 1 - P(C \leq 49).$

Steps:

1. Using the binomial cumulative probability $P(C \leq k)$: $P(C \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}.$ Direct computation is tedious, so we use an approximation.

2. Normal Approximation to the Binomial Distribution: For large $n$, the binomial distribution $\mathcal{B}(n, p)$ can be approximated by a normal distribution: $C \sim \mathcal{N}(\mu, \sigma^2),$ where: $\mu = n \cdot p = 60 \cdot 0.75 = 45,$ $\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{60 \cdot 0.75 \cdot 0.25} = \sqrt{11.25} \approx 3.354.$

3. Convert $P(C \geq 50)$ to the standard normal form: $P(C \geq 50) \approx P\left(Z \geq \frac{50 - \mu}{\sigma}\right),$ where $Z$ is the standard normal variable: $Z \geq \frac{50 - 45}{3.354} \approx \frac{5}{3.354} \approx 1.49.$

4. From standard normal tables, $P(Z \geq 1.49)$: $P(Z \geq 1.49) = 1 - P(Z \leq 1.49) \approx 1 - 0.9319 = 0.0681.$

Thus, the probability $P(C \geq 50)$ is approximately 0.068, rounded to the nearest thousandth.

Summary of Results

1. $C \sim \mathcal{B}(60, 0.75)$.
2. $P(C \geq 50) \approx 0.068$.

Do you need further explanation or more steps elaborated? Here are five related questions for deeper understanding:

1. How does the normal approximation simplify binomial calculations?
2. What would happen to $\sigma$ if the probability of success were $0.5$ instead of $0.75$?
3. How can this method be applied to real-life problems involving probabilities?
4. What are the limitations of using the normal approximation for small $n$?
5. How would the probability change if we calculated $P(C = 50)$ instead of $P(C \geq 50)$?

Tip: Always check the conditions for using approximations, especially when $n$ is small or $p$ is close to 0 or