# Probability of a random variable

• Shackman
In summary, the conversation discusses a problem involving 5 men and 5 women ranked according to exam scores. The task is to find the probability of the highest ranking being achieved by a woman. Through a series of calculations, it is found that the probabilities for X=1 to X=6 are 50%, 27.78%, 13.9%, 5.95%, 1.98%, and 0.397%, respectively. These probabilities add up to 100.07%.
Shackman

## Homework Statement

5 men and 5 women are ranked according to exam scores. Assume no two scores are the same and each 10! rankings are equally likely. Let random variable X denote the highest ranking achieved by a woman e.g. X=2 means the highest test score was achieved by 1 of the 5 men and the second highest by 1 of the 5 women. Find P{X=i}, i=1,2,3,...,10.

## The Attempt at a Solution

I am quite lost on this problem, I have a very poor background in statistics so I'm not sure if there is a specific formula I should be using or what the most efficient way of doing this is, but P{X=i}, i=7,8,9,10 is 0 because there are only 5 men so the lowest a woman can be ranked is 6th place. I think P(X=1) = 5* 9!/10!, P(X=2)=5*8!/10!, P(X=3)=5*7!/10!, P(X=4)=5*6!/10!, P(X=5)=5*5!/10! and P(X=6)=5*4!/10! since there are 5*9! different rankings (9! combinations times 5 different possible women ranked at that position) with a woman ranked first and so on. So is P(X=i), i=1,2,...,10 the summation of these probabilities or P(X=i)=.567 (assuming they are correct)?

Last edited:
Hi Shackman,

P(X=1): Choose a woman and place her at the top (there are nCr(5,1)=5 choices possible here). Then there are 9! permutations of the remaining people, for a total of 5*9! configurations.

P(X=2): Choose one male (again 5 choices here), place him at the top, and one woman (5 choices again) and place her at second. There are then 8! permutations of the remaining people, for a total of 5^2*8! combinations.

So it looks like you were on the right track for the first few. Repeat this type of procedure for the rest choosing (X-1) men and placing them above the woman, choosing one woman and placing her at the X spot, then permuting the leftovers. However, you must take into account that you must also permute the number of men that you place before the women!

If you want to check, the probabilities I get are

X=1: 50%
X=2: 27.78%
X=3: 13.9%
X=4: 5.95%
X=5: 1.98%
X=6: 0.397%
---------
Total probability: 100.07% (I'm off due to rounding...)

## What is the definition of probability of a random variable?

The probability of a random variable is the likelihood or chance that a particular outcome will occur when the variable is measured or observed. It is a numerical representation of uncertainty.

## How is the probability of a random variable calculated?

The probability of a random variable is calculated by dividing the number of outcomes that satisfy a certain condition by the total number of possible outcomes. This is known as the classical definition of probability.

## What is the difference between discrete and continuous random variables?

Discrete random variables take on a finite or countable number of values, while continuous random variables can take on any value within a certain range. For example, the number of children in a family is a discrete random variable, while the height of a person is a continuous random variable.

## How is the probability of a random variable represented?

The probability of a random variable can be represented in different ways, such as a fraction, decimal, or percentage. It can also be represented graphically using a probability distribution, such as a bar graph or a probability density function.

## Why is understanding the probability of a random variable important?

Understanding the probability of a random variable is important in many areas of science, including statistics, economics, and physics. It allows us to make predictions and make informed decisions based on data and uncertainty. It also helps us to understand the likelihood of certain events occurring and the potential risks associated with them.

• Calculus and Beyond Homework Help
Replies
3
Views
810
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
380
• Calculus and Beyond Homework Help
Replies
13
Views
1K
• Calculus and Beyond Homework Help
Replies
19
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
707
• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
988
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
11
Views
1K