Answer :
Sure! Let's solve this problem step-by-step.
### Step 1: Understanding the Problem
We have the following parameters:
- The probability of success, \( p = 0.61 \), which means there is a 61% chance that a human resource manager says job applicants should follow up within two weeks.
- The number of trials, \( n = 20 \), which indicates that 20 human resource managers are randomly selected.
- We are interested in the case where exactly \( k = 14 \) managers say that applicants should follow up within two weeks.
### Step 2: Binomial Distribution
We use the binomial distribution to solve this problem. The binomial distribution formula for finding the probability of exactly \( k \) successes in \( n \) independent trials is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient,
- \( p^k \) is the probability of having \( k \) successes,
- \( (1-p)^{n-k} \) is the probability of having \( n-k \) failures.
### Step 3: Plugging in the Values
Now let's plug in the values into the formula:
- \( n = 20 \)
- \( k = 14 \)
- \( p = 0.61 \)
- \( 1 - p = 0.39 \)
First, we calculate the binomial coefficient:
[tex]\[ \binom{20}{14} = \frac{20!}{14!(20-14)!} = \frac{20!}{14! \cdot 6!} \][/tex]
Next, we calculate \( p^k \) and \( (1 - p)^{n - k} \):
[tex]\[ p^{14} = 0.61^{14} \][/tex]
[tex]\[ (1 - p)^{6} = 0.39^{6} \][/tex]
### Step 4: Computing the Probability
Multiplying these components together gives us the probability \( P(X = 14) \).
After performing the calculations (simplified here for clarity), we find:
[tex]\[ P(X = 14) \approx 0.1347 \][/tex]
### Step 5: Conclusion
Thus, the probability that exactly 14 out of 20 randomly selected human resource managers say job applicants should follow up within two weeks is approximately \( 0.1347 \).
### Final Answer
The probability is [tex]\( \boxed{0.1347} \)[/tex].
### Step 1: Understanding the Problem
We have the following parameters:
- The probability of success, \( p = 0.61 \), which means there is a 61% chance that a human resource manager says job applicants should follow up within two weeks.
- The number of trials, \( n = 20 \), which indicates that 20 human resource managers are randomly selected.
- We are interested in the case where exactly \( k = 14 \) managers say that applicants should follow up within two weeks.
### Step 2: Binomial Distribution
We use the binomial distribution to solve this problem. The binomial distribution formula for finding the probability of exactly \( k \) successes in \( n \) independent trials is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient,
- \( p^k \) is the probability of having \( k \) successes,
- \( (1-p)^{n-k} \) is the probability of having \( n-k \) failures.
### Step 3: Plugging in the Values
Now let's plug in the values into the formula:
- \( n = 20 \)
- \( k = 14 \)
- \( p = 0.61 \)
- \( 1 - p = 0.39 \)
First, we calculate the binomial coefficient:
[tex]\[ \binom{20}{14} = \frac{20!}{14!(20-14)!} = \frac{20!}{14! \cdot 6!} \][/tex]
Next, we calculate \( p^k \) and \( (1 - p)^{n - k} \):
[tex]\[ p^{14} = 0.61^{14} \][/tex]
[tex]\[ (1 - p)^{6} = 0.39^{6} \][/tex]
### Step 4: Computing the Probability
Multiplying these components together gives us the probability \( P(X = 14) \).
After performing the calculations (simplified here for clarity), we find:
[tex]\[ P(X = 14) \approx 0.1347 \][/tex]
### Step 5: Conclusion
Thus, the probability that exactly 14 out of 20 randomly selected human resource managers say job applicants should follow up within two weeks is approximately \( 0.1347 \).
### Final Answer
The probability is [tex]\( \boxed{0.1347} \)[/tex].
