This problem involves a binomial probability calculation. Here's the breakdown:

Step 1: Define the binomial probability formula

The probability of $X$ successes in $n$ trials is given by: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ where:

• $n = 30$ (number of trials),
• $k = 21$ (number of successes),
• $p = 0.46$ (probability of success in a single trial),
• $1-p = 0.54$ (probability of failure in a single trial),
• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the number of combinations.

Step 2: Plug in the values

$P(X = 21) = \binom{30}{21} (0.46)^{21} (0.54)^9$

Step 3: Calculate each component

1. Combinations: $\binom{30}{21} = \frac{30!}{21!(30-21)!} = \frac{30!}{21! \cdot 9!} = 14307150$

2. Success probability raised to the power: $(0.46)^{21} \approx 6.191 \times 10^{-8}$

3. Failure probability raised to the power: $(0.54)^9 \approx 0.00298$

Step 4: Multiply these values

$P(X = 21) = 14307150 \cdot (6.191 \times 10^{-8}) \cdot 0.00298$

$P(X = 21) \approx 0.0264$

Final Answer:

The probability that exactly 21 of the 30 human resource managers say job applicants should follow up within two weeks is approximately 0.0264 or 2.64%.

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Would you like further clarification or an explanation of any step? Here are some related questions to expand on:

1. How does the binomial distribution formula generalize to other probabilities?
2. What if the problem asked for at least 21 managers? How would that change the calculation?
3. How does increasing the sample size (e.g., from 30 to 50 managers) affect this probability?
4. Can this problem be solved using a normal approximation to the binomial distribution?
5. How would the probability change if the success rate $p$ were higher or lower?

Tip: When solving binomial problems, double-check the factorial terms in the combinations to avoid computational errors!