This problem involves a binomial probability distribution because we are dealing with a fixed number of trials ($n = 8$), two possible outcomes (success or failure), and a constant probability of success ($p = 0.51$).

We are tasked with finding the probability that at least 2 of the 8 randomly selected human resource managers say job applicants should follow up within two weeks. To calculate this, we need to sum the probabilities for $x = 2, 3, \dots, 8$.

Complementary Rule

Instead of calculating $P(X \geq 2)$ directly, we use the complement rule: $P(X \geq 2) = 1 - P(X < 2)$ Where $P(X < 2) = P(X = 0) + P(X = 1)$.

Binomial Probability Formula

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

• $\binom{n}{x} = \frac{n!}{x!(n-x)!}$ is the number of combinations,
• $p = 0.51$ is the probability of success,
• $1-p = 0.49$ is the probability of failure.

Step-by-Step Calculation

1. Calculate $P(X = 0)$: $P(X = 0) = \binom{8}{0} (0.51)^0 (0.49)^8 = 1 \cdot (1) \cdot (0.49)^8 \approx 0.00192$

2. Calculate $P(X = 1)$: $P(X = 1) = \binom{8}{1} (0.51)^1 (0.49)^7 = 8 \cdot (0.51) \cdot (0.49)^7 \approx 0.01528$

3. Sum $P(X = 0)$ and $P(X = 1)$: $P(X < 2) = P(X = 0) + P(X = 1) \approx 0.00192 + 0.01528 = 0.0172$

4. Find $P(X \geq 2)$: $P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0172 = 0.9828$

Final Answer:

The probability that at least 2 of the 8 human resource managers say job applicants should follow up within two weeks is approximately: $\boxed{0.9828}$

Would you like further explanation or assistance with related problems? Here are 5 related questions:

1. How do we calculate probabilities for exactly 3 successes in a binomial distribution?
2. What changes if we increase the sample size from 8 to 10 human resource managers?
3. How would this problem change if the probability of success were different, e.g., $p = 0.6$?
4. Can we approximate this binomial problem using a normal distribution?
5. What are some real-world applications of binomial distributions in HR?

Tip: When dealing with "at least" probabilities, using the complement rule often simplifies the calculation!