I really could not find a math forum... but since I am an accounting major I'd figure I'd ask here. I am taking a statistics course and we just started probabilities. I am terrible at that and I need help...

Suppose that 18% of the employees at a given corporation engage in physical activity during lunch. Moreover, assume that 57% of all employees are male, and 12% of all employees are males who engage in physical activity during lunch.

a. If chosen at random, what is the probability that person is female who engages in physical activity during lunch?

b. if chose at random, what is probability that person is female and does not engage in physical activity during lunch?


thanks in advance!

a) 6%
b) 37%

Think of the company as 100 employees. If 18% do the activity this mean 18 employees. If 12 males of the 100 employees do the activity, then 6 females must make up the remainder of the 18. Furthermore, if 57 employees are male, then 43 are female (necessary to arrive at 100, or 100%). Since 6 of the females do the activity, this leaves 37 that do not.

6/100 = Females who do physical activity
37/100 = Females who do not do the activity

Total female population = 6 + 37 = 43

Total populations = 100 = 57 males + 43 females