Question

a large company states in its promotional literature that 74% of its employees have college degrees. assume this claim is true. if 4 employees are selected at random from this company, what is the probability that at least 1 of the selected employees has a college degree? 0.0046 0.2999 0.7001 0.9954

Explanation:

Step1: Calculate the probability of an employee not having a college degree

The probability that an employee has a college degree \(p = 0.74\). So the probability that an employee does not have a college degree \(q=1 - p=1 - 0.74 = 0.26\)

Step2: Calculate the probability that none of the 4 employees have a college degree

Using the binomial probability formula \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(n = 4\), \(k = 0\), \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(p = 0.74\), \(q = 0.26\)

When \(k = 0\), \(C(4,0)=\frac{4!}{0!(4-0)!}=1\)

\(P(X = 0)=C(4,0)\times(0.74)^{0}\times(0.26)^{4}=1\times1\times0.26^{4}\)

\(0.26^{4}=0.26\times0.26\times0.26\times0.26 = 0.00456976\)

Step3: Calculate the probability that at least 1 has a college degree

The probability that at least 1 has a college degree \(P(X\geq1)=1 - P(X = 0)\)

\(P(X\geq1)=1-0.00456976 = 0.99543024\approx0.9954\)

Answer:

0.9954