We can solve this problem using the binomial probability formula. The binomial distribution describes the number of successes in a fixed number of independent trials, where each trial has two possible outcomes (success or failure). In this case, success is defined as passing the test.

The binomial probability formula is:

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

Where:

• $P(X = k)$ is the probability of exactly $k$ successes.
• $n$ is the number of trials (in this case, 12 applicants).
• $k$ is the number of successes (7 applicants passing the test).
• $p$ is the probability of success on each trial (65% = 0.65).
• $\binom{n}{k}$ is the binomial coefficient, or "n choose k," which is calculated as:

$\binom{n}{k} = \frac{n!}{k!(n - k)!}$

Now, let’s calculate:

1. Given data:

   • $n = 12$
   • $k = 7$
   • $p = 0.65$

2. Binomial coefficient:

$\binom{12}{7} = \frac{12!}{7!(12 - 7)!} = \frac{12!}{7!5!} = 792$

3. Probability calculation:

$P(X = 7) = 792 \cdot (0.65)^7 \cdot (0.35)^5$

Let me calculate the final probability:

$P(X = 7) \approx 792 \cdot 0.04915 \cdot 0.00525$ $P(X = 7) \approx 792 \cdot 0.00025795 \approx 0.2042$

So, the probability that exactly 7 applicants will pass the test is approximately 0.2042, or 20.42%.

Would you like further details or have any questions about this calculation?

Here are 5 related questions:

1. What is the probability that fewer than 7 applicants will pass the test?
2. What is the expected number of applicants who will pass the test?
3. How would the probability change if 15 applicants were tested instead of 12?
4. What is the probability that at least 9 applicants will pass?
5. Can you explain how the binomial distribution applies to real-life scenarios?

Tip: In a binomial distribution, as the number of trials (n) increases, the distribution starts resembling a normal distribution under certain conditions (central limit theorem).