To solve this problem, we'll need to use the binomial probability formula. The binomial probability formula is used to find the probability of a specific number of successes in a certain number of trials and is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\( n \)[/tex] is the number of trials (in this case, 14 human resource managers).
- [tex]\( k \)[/tex] is the number of successes (in this case, fewer than 3 managers).
- [tex]\( p \)[/tex] is the probability of success on a single trial (in this case, 0.55).
- [tex]\(\binom{n}{k}\)[/tex] is the combination of [tex]\( n \)[/tex] items taken [tex]\( k \)[/tex] at a time, calculated as [tex]\(\frac{n!}{k!(n-k)!}\)[/tex].
To find the probability that fewer than 3 managers say job applicants should follow up within two weeks, we need to calculate the probability for each scenario where the number of managers is 0, 1, or 2 and then sum those probabilities.
1. Probability for exactly 0 managers ([tex]\( P(X = 0) \)[/tex]):
[tex]\[
P(X = 0) = \binom{14}{0} (0.55)^0 (0.45)^{14} = 1 \cdot 1 \cdot (0.45)^{14}
\][/tex]
2. Probability for exactly 1 manager ([tex]\( P(X = 1) \)[/tex]):
[tex]\[
P(X = 1) = \binom{14}{1} (0.55)^1 (0.45)^{13} = 14 \cdot (0.55) \cdot (0.45)^{13}
\][/tex]
3. Probability for exactly 2 managers ([tex]\( P(X = 2) \)[/tex]):
[tex]\[
P(X = 2) = \binom{14}{2} (0.55)^2 (0.45)^{12} = 91 \cdot (0.55)^2 \cdot (0.45)^{12}
\][/tex]
Next, add up these probabilities:
[tex]\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\][/tex]
According to the numerical calculations, the sum of these probabilities is given as approximately 0.0022.
Therefore, the probability that fewer than 3 of the 14 human resource managers say job applicants should follow up within two weeks is approximately:
[tex]\[ \boxed{0.0022} \][/tex]
Rounded to four decimal places as required, the probability is:
[tex]\[ 0.0022 \][/tex]
