Probability Distributions
Probability distributions show the probabilities of all possible outcomes of an experiment. The sum of the probabilities in any distribution is always 1.
In probability experiments, the outcomes are often numerical, meaning they can be counted or measured. A Random Variable is used to represent these outcomes in numerical form.
Random Variable
A random variable, \(X\), has as single value(denoted \(x\) for each outcome in the experiment. For example, if \(X\) is the number rolled with a die, then \(x\) has a different value for each of the six possible outcomes.
There are two Different Types of random variables:
- Discrete Distribution: This type of distribution consists of whole numbers. An example of a Discrete Distribution is the number of heads obtained when flipping a coin multiple times
- Continuous Distribution: This type of distribution can take any value within a range, including decimals. An example of a Continuous Distribution is the time taken to finish a race
The probability of getting a specific value for a random variable is \(P(X = x)\) or just \(P(x)\).
Non-Random Probability Distribution
A non-random probability distribution typically doesn’t exist since probability distributions describe random variables with uncertain outcomes. Non-random variables have fixed values, so their "distribution" would assign a probability of 1 to a single outcome.
Example
A fair die is rolled once. What is the probability of rolling a 4?
The possible outcomes of rolling a die are 1, 2, 3, 4, 5, and 6. Since the die is fair, each outcome has an equal chance. The probability of rolling a 4 is:
Therefore, we can determine the probability of rolling a 4 is \(\mathbf{\cfrac{1}{6}}\).
For Continuous Distributions, the probability of getting an exact value is 0 because there are infinite possible values in the range. So, the probability is:
Therefore, the probability of the height being exactly 170cm is 0.
Uniform Distribution
Uniform distribution is a type of probability distribution can be classified as either continuous or discrete, depending on the nature of the outcome. In a uniform distribution, all outcomes are equally likely, meaning each result has the same probability of occurring.
In a discrete uniform distribution, if there are \(n\) possible outcomes, then the probability of each outcome is given by the following formula:
Where \(n\) represents the total number of possible outcomes.
Expectation and Expected Value
Expectation and expected value represent the average outcome of a probability distribution. The expected value is calculated by multiplying each outcome by its probability and summing the results.
This can be represented algebraically as such:
Where:
- \( x_1, x_2, x_3, \dots, x_n \) are the possible outcomes (Random Variables)
- \( P(x_1), P(x_2), \dots, P(x_n) \) are the probabilities of each outcome
A nature retreat has 7 4.5-meter kayaks, 4 5.1-meter kayaks, 2 5.3-meter kayaks, and 5 6.0-meter kayaks. Kayaks are randomly assigned to adventurers going on a river trip.
- Show the Probability Distribution for the length of an assigned kayak
- Calculate the Expected Value for the length of an assigned kayak
i. First, we can determine the total number of kayaks.
Next, we can determine the probability of each kayak length:
\(P(4.5) = \cfrac{7}{18}\)
\(\quad P(5.1) = \cfrac{4}{18}\)
\(\quad P(5.3) = \cfrac{2}{18}\)
\(\quad P(6.0) = \cfrac{5}{18}\)
Therefore, we can determine that the probabilities of the 4.5, 5.1, 5.3, and 6.0-meter kayaks are \(\mathbf{\cfrac{7}{18}}\), \(\mathbf{\cfrac{4}{18}}\), \(\mathbf{\cfrac{2}{18}}\), and \(\mathbf{\cfrac{5}{18}}\) respectively.
ii. In order to determine the Expected Value of the assigned kayak, we can multiply the kayak lenghts by their respective probabilities and find the sum of the values:
\(E(X) = \left(4.5 \times \cfrac{7}{18}\right) + \left(5.1 \times \cfrac{4}{18}\right) + \left(5.3 \times \cfrac{2}{18}\right) + \left(6.0 \times \cfrac{5}{18}\right)\)
\(E(X) = \cfrac{31.5}{18} + \cfrac{20.4}{18} + \cfrac{10.6}{18} + \cfrac{30}{18}\)
\(E(X) = \cfrac{92.5}{18} \approx 5.2 \)
Therefore, we can determine the Expected Length of the kayak is 5.2 meters.