A town recently dismissed 9 employees in order to meet their new budget reductions. The town had 7 employees over 50 years of age and 15 under 50. If the dismissed employees were selected at random, what is the probability that exactly 1 employee was over 50? Express your answer as a fraction or a decimal number rounded to four decimal places.

[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of x successes out of n trials, p is the probability of success on a particular trial.

Given : The number of employees over 50 years of age =7

The probability of employees over 50 years of age = [tex]\dfrac{7}{7+15}=0.318[/tex]

Number of dismissed employees : n= 9

Now, the required probability will be :

[tex]P(x =1)=^{9}C_1(0.318)^{1}(1-0.318)^{9-1}\\\\=(9)(0.318)(1-0.318)^{8}=0.133950588714\approx0.1340[/tex]

Thus, the probability that exactly 1 employee was over 50 = 0.1340