60% of a firm's employees are men. Suppose four of the firm's employees are selected. a. What is the random variable we can consider here? b. What is probability of finding three men and one woman? c. What is the answer for part if 75% of employees were men?

60% of the employees are men.

Then the 40% left must be women.

4 of them are selected.

a)  What is the random variable we can consider here?

A random variable is a variable that can have different outcomes or values, in other words, it is a stochastic variable.

We have a selection of 4 employees at random, then we can define the variable X, that represents (for each employee) the possibility of being man or woman.

b) Probability of 3 men and 1 woman.

Suppose that the selection is in the order:

man-man-man-woman.

We have:

0.6 tree times.

0.4 one time.

P = 0.6^3*0.4

But this is for the case where the woman is in the last selection, it can be in the first, second and third, then we have 4 permutations, the actual probability is:

P = 4*( 0.6^3*0.4) = 0.3456.

c) If now we have tat 75% are men, then 25% are women.

Now we can do the same as before, but now the probability for a man is 0.75 and for a woman is 0.25, then the previous equation is now:

P = 4*(0.75^3*0.25) = 0.421875

the mean and mad

step-by-step explanation:

idk bring the function back   im not smart

step-by-step explanation: