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Part 1
Assume that when human resource managers are randomly​ selected, 57​% say job applicants should follow up within two weeks. If 6 human resource managers are randomly​ selected, find the probability that at least 2 of them say job applicants should follow up within two weeks.
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Part 1
The probability is 
  
enter your response here.
​(Round to four decimal places as​ needed.)

This problem can be solved using the **binomial probability formula**, which is used when there are only two outcomes: success or failure. In this case, success is defined as an HR manager saying that job applicants should follow up within two weeks, and failure is the opposite.

### Step-by-Step Solution

#### Step 1: Define the parameters for the binomial distribution.

- The probability of success, \( p = 0.57 \) (57%).
- The probability of failure, \( q = 1 - p = 0.43 \).
- The number of trials (HR managers), \( n = 6 \).
- The minimum number of successes (at least 2 managers say follow up), \( x \geq 2 \).

#### Step 2: Use the complement rule to find the probability.
It’s easier to calculate the probability of fewer than 2 successes, and then subtract from 1 to find the probability of at least 2 successes:

\[
P(\text{at least 2 successes}) = 1 - P(\text{0 or 1 success})
\]

This means we need to find the probabilities for:
1. 0 successes
2. 1 success

#### Step 3: Use the binomial probability formula.

The binomial probability formula is:

\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]

Where:
- \( \binom{n}{k} \) is the binomial coefficient (combinations),
- \( p \) is the probability of success,
- \( q \) is the probability of failure,
- \( n \) is the number of trials,
- \( k \) is the number of successes.

Now calculate for:
1. \( P(X = 0) \)
2. \( P(X = 1) \)

#### Step 4: Calculate probabilities.

Let's calculate \( P(X = 0) \) and \( P(X = 1) \), then find \( P(X \geq 2) \).

I'll compute these probabilities now.The probability that at least 2 of the 6 randomly selected human resource managers say job applicants should follow up within two weeks is **0.9434**.

Would you like further details or have any other questions?

Here are 5 related questions to explore:
1. What is the probability that exactly 2 managers say follow up within two weeks?
2. How would the result change if the probability of success were different, say 70%?
3. What is the probability that none of the managers recommend following up?
4. How is the binomial distribution different from a normal distribution?
5. How can this method be applied to larger populations, like 100 HR managers?

**Tip**: The complement rule is often useful when it's easier to calculate the opposite scenario.

Assume that when human resource managers are randomly selected, 57% say job applicants should follow up within two weeks. If 6 human resource managers are randomly selected, find the probability that at least 2 of them say job applicants should follow up within two weeks. (Round to four decimal places as needed.)

Learn how to calculate the probability that at least 2 out of 6 human resource managers say job applicants should follow up within two weeks. This problem involves the binomial distribution formula and complement rule, making it suitable for Grades 10-12.

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