This problem can be solved using the binomial probability formula, which is used when you have a fixed number of independent trials, each with two possible outcomes (success or failure). Here, passing the test is considered a "success."

The binomial probability formula is:

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

Where:

• $n = 12$ (the number of applicants),
• $k = 7$ (the number of applicants who pass),
• $p = 0.65$ (the probability that a single applicant passes),
• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

Step-by-step solution:

1. Set up the formula: $P(X = 7) = \binom{12}{7} (0.65)^7 (0.35)^5$

2. Find the binomial coefficient $\binom{12}{7}$: $\binom{12}{7} = \frac{12!}{7!(12-7)!} = \frac{12!}{7!5!} = 792$

3. Calculate the powers of the probabilities: $(0.65)^7 = 0.0493 \quad \text{and} \quad (0.35)^5 = 0.00525$

4. Multiply all the components together: $P(X = 7) = 792 \times 0.0493 \times 0.00525 = 0.205$

Thus, the probability that exactly 7 out of 12 applicants will pass the test is approximately 0.205, or 20.5%.

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Would you like any further details or have any questions?

Here are some related questions:

1. What is the probability that more than 7 applicants will pass?
2. How would the probability change if the passing rate were 70% instead of 65%?
3. What is the probability that fewer than 7 applicants will pass?
4. How does the binomial distribution change with different sample sizes?
5. What is the expected number of applicants to pass out of 12?

Tip: In binomial problems, if you're dealing with large sample sizes, you can use the normal approximation to the binomial for faster computations.