A fitness centre has 16 employees. One of the 16 employees is selected at random. The probability that it is a woman working part time is $1/8$. Two of the 16 employees are selected at random. The probability that they are both men working full time is $1/8$. Complete the table of information below about the 16 employees of the fitness centre. Show all supporting calculations clearly. Answer

1. Find the number of women working part time.

Probability that a randomly selected employee is a woman working part time is $\frac18$.

Since there are 16 employees,
$\text{number of women working part time}=\frac18\times 16=2.$

2. Let the number of men working full time be an unknown and form an equation using probability without replacement.

Let the number of men working full time be $x$.

Two employees are selected at random without replacement. Hence
$\frac{x}{16}\times \frac{x-1}{15}=\frac18.$

3. Solve for the number of men working full time.

$\frac{x(x-1)}{16\times 15}=\frac18$
$x(x-1)=30$
$x^2-x-30=0$
$(x-6)(x+5)=0$

So $x=6$ or $x=-5$.

Since the number of employees cannot be negative, $x=6$.

4. Complete the remaining table entry using the total number of employees.

Known entries:

Women part time $=2$, women full time $=5$, men full time $=6$.

Total accounted for:
$2+5+6=13.$

Therefore men part time:
$16-13=3.$