A certain company has n+m employees, consisting of n women...

A certain company has $n+m$ employees, consisting of $n$ women and $m$ men. The company is deciding which employees to promote. (a) Suppose for this part that the company decides to promote $t$ employees, where $1 \leq t \leq n+m$, by choosing $t$ random employees (with equal probabilities for each set of $t$ employees). What is the distribution of the number of women who get promoted? (b) Now suppose that instead of having a predetermined number of promotions to give, the company decides independently for each employee, promoting the employee with probability $p$. Find the distributions of the number of women who are promoted, the number of women who are not promoted, and the number of employees who are promoted. (c) In the set-up from (b), find the conditional distribution of the number of women who are promoted, given that exactly $t$ employees are promoted.

A certain company has $n+m$ employees, consisting of $n$ women and $m$ men. The company is deciding which employees to promote.
(a) Suppose for this part that the company decides to promote $t$ employees, where $1 \leq t \leq n+m$, by choosing $t$ random employees (with equal probabilities for each set of $t$ employees). What is the distribution of the number of women who get promoted?
(b) Now suppose that instead of having a predetermined number of promotions to give, the company decides independently for each employee, promoting the employee with probability $p$. Find the distributions of the number of women who are promoted, the number of women who are not promoted, and the number of employees who are promoted.
(c) In the set-up from (b), find the conditional distribution of the number of women who are promoted, given that exactly $t$ employees are promoted.

Key Concepts

Conditional Probability Distribution

A conditional probability distribution describes the probability of an event occurring given that another event has already occurred. In scenarios where the total number of successes is fixed (for example, t employees promoted), the distribution of subcategories (such as the number of promoted women) must be adjusted according to this condition. This often results in a hypergeometric form when the conditioned event influences the remaining arms of the experiment.

Binomial Distribution

The binomial distribution applies when there is a fixed number of independent and identical trials, each resulting in a success or failure with constant probability. In contexts where each individual decision (such as promotion) is made independently with the same probability of success, the number of successes in the group follows a binomial distribution. It is defined by the total number of trials and the fixed probability of success in each trial.

Hypergeometric Distribution

This distribution arises when sampling without replacement from a finite population containing two types of items. It is used to model the probability of drawing a specific number of 'successes' (for example, promoted women) when a fixed number of items are chosen at random from a larger group. The key idea is that the probabilities change with each draw because the population composition changes, which distinguishes it from distributions based on independent trials.

Transcript

00:01 The telechatronic company hired eight employees from a large pool of applicants with an equal number of males and females.

00:08 If the hiring is done without regard to sex, the number of females hired, and the corresponding probabilities are as follows.

00:14 So i put it into a table format.

00:17 Does this given information describe a probability distribution? so yes, our probabilities add up to 0 .999, which is an acceptable.

00:31 Amount and none of our probabilities are greater than one or less than zero so we can proceed.

00:40 Part b, assuming that a probability distribution is described, find its mean and standard deviation.

00:46 This is where statistical software comes in handy because to find the mean, i need to take x times p of x, and then i need to find the sum of those in order to find my mean.

01:00 So i'm going to to use my graphing calculator and i'm going to enter my x column into list one and my probability into list two because then i can tell my graphing calculator to take list one times list two and my calculator will propagate all of those products for me.

01:21 I can then ask my calculator to do a one variable statistic and it will find the sum for me.

01:27 So the sum using my statistical software, was 3 .996.

01:37 For the standard deviation, again, statistical software is a lifesaver because by hand i would need to take x minus my newly found mu.

01:48 And then i would square that and then multiply at times my probability of x.

01:55 But with statistical software, what i would do is i would say, parentheses, list 1 minus 3 .996.

02:05 Paranthesis squared times list 2...

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