This problem can be solved using the geometric probability distribution, which models the probability of the first success occurring on a specific trial. The probability of Carl passing the test on any given attempt is $p = 0.30$, and the probability of failing is $1 - p = 0.70$.

We are asked to find the probability that Carl passes between his second and fourth attempt, meaning he passes on the 2nd, 3rd, or 4th attempt.

Steps to solve:

1. Geometric Probability Formula: The probability of Carl passing the test on the $n$-th attempt is given by: $P(X = n) = (1 - p)^{n-1} \cdot p$ where $X$ is the trial number of the first success.

2. Find the probability of passing on the 2nd, 3rd, and 4th attempts:

   • Probability of passing on the 2nd attempt: $P(X = 2) = (0.70)^1 \cdot 0.30 = 0.21$
   • Probability of passing on the 3rd attempt: $P(X = 3) = (0.70)^2 \cdot 0.30 = 0.147$
   • Probability of passing on the 4th attempt: $P(X = 4) = (0.70)^3 \cdot 0.30 = 0.1029$

3. Sum the probabilities: The total probability that Carl passes between his 2nd and 4th attempt is the sum of the probabilities of passing on the 2nd, 3rd, and 4th attempts: $P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)$ $P(2 \leq X \leq 4) = 0.21 + 0.147 + 0.1029 = 0.4599$

Thus, the probability that Carl passes the test between his second and fourth attempt is 0.4599.

Final Answer:

The correct option is 0.4599.

Would you like more details or have any questions?

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Here are five related questions you might find interesting:

1. What is the probability that Carl passes the test on his first attempt?
2. How would the probabilities change if Carl had a 50% chance of passing each attempt?
3. What is the expected number of attempts Carl would need to pass?
4. How can we calculate the probability that Carl passes on his 5th attempt?
5. If Carl only had 3 attempts, what is the probability he passes at least once?

Tip: When dealing with geometric distributions, always focus on the independence of each trial. Each attempt’s probability is unaffected by previous attempts.