Embedded Files
1. Concepts & Definitions
1.1. Experiment, observation, and sample space
1.2. Sample space: Venn and Tree diagram
1.3. Simple and composite events
1.4. Three definitions of probability
1.5. Law of large numbers and its consequences
1.6. Frequency and empirical probability
2. Problem & Solution
2.4. Frequency of categories from tables
2.5. Simple and marginal probabilities
2.6. Conditional probabilities
Factorial
Factorial
The symbol n! mean n factorial, and represents the product of all the whole numbers from 1 to n. That is:
n! = n*(n 1)* ... *2*1
0! = 1
Furthermore:
n! = n*(n 1)!
Example 1: 5! = 5*4*3*2*1 = 120
Example 2: (5-5)! = 0! = 1
Example 3: 6! = 6*5! = 6*120 = 720
Combinations
Combinations
Gives the number of ways that a set of n elements can be combined to form subsets containing x elements.
nCx = Cn,x = n!/x!(n-x)!
Three options of coffee capsules will be chosen from a total of five (A, B, C, D, and E). How many are possible combinations?
5C3 = 5!/3!(5-3)! = 10
The next figure illustrates how to make combinations for this example.
Use combinations to compute probabilities
Use combinations to compute probabilities
In the previous examples, we calculated the number of combinations for different scenarios. However, we didn’t use them to calculate probabilities. In this context, probability is the number of combinations considered to be an event divided by the total number of combinations [1].
The classic example of probability using combinations without repetition is a lottery where machines randomly choose balls with numbers from a pool of balls. The order of the numbers does not matter. You just need to match the numbers. Because each ball can be drawn only once, it is without repetition. For this example, we’ll look at the Powerball lottery, which is a lottery offered in 45 states in the United States. In the basic game, there are 69 white balls in a machine that randomly selects five of the balls.
The following ratio defines the probability of winning: Probability formula. The numerator is simple. There is one winning combination without repetition. You could use the equation to calculate 5C5, which equals the number of combinations where you have five numbers in the winning combination, and there are five winning numbers from which to choose. But it makes sense intuitively. There’s just one winning combination. For the denominator, you need to calculate 69C5, which equals the number of combinations when you draw five numbers from a total of 69 numbers. Let’s enter these numbers into the equation: 69C5 = 11,238,513.
When you draw five numbers out of 69 without repetition, there are 11,238,513 combinations. Using the standard notation, the probability of winning the basic game in Powerball is the following: You have a 1 in 11,238,513 chance of winning the basic game in the Powerball lottery.
Permutation with repetition
Permutation with repetition
When the outcomes in a permutation can repeat, statisticians refer to it as permutations with repetition. For example, in a four-digit PIN, you can repeat values, such as 1-1-1-1. Analysts also call this permutation with replacement. To calculate the number of permutations, take the number of possibilities for each event and then multiply that number by itself X times, where X equals the number of events in the sequence. For example, with four-digit PINs, each digit can range from 0 to 9, giving us 10 possibilities for each digit. We have four digits. Consequently, the number of permutations with repetition for these PINs = 10 * 10 * 10 * 10 = 10,000. We write this mathematically as[2]:
nr
Where: n = the number of possible outcomes for each event.
For instance, n = 10 for the PIN example. r = the size of each permutation. For example, r = 4 for a four-digit pin.
Permutation without repetition
Permutation without repetition
When the outcomes cannot be repeated, statisticians call them permutations without repetition. This situation frequently occurs when you’re working with unique physical objects that can occur only once in a permutation. Imagine you have 10 different books and want to calculate how many possible ways you can arrange them on a bookshelf. After you place the first book, the second book must be a different book. Consequently, this is an example of permutations without repetition. Analysts also call this permutation without replacement. Mathematically, we’d calculate the permutations for the book example using the following method [2]:
n! = 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800 There are 3,628,800 permutations for ordering 10 books on a shelf without repeating books.
Partial Permutation without repetition
Partial Permutation without repetition
In some cases, you want to consider only a portion of the possible permutations. In the bookshelf example, we wanted to know the total number of 10 books. But what if we could fit only five of the 10 books on the shelf? How many permutations of five books are possible using our 10 books? Use the following formula to calculate the number of arrangements of r items from n objects. There are several standard methods that statisticians use to notate permutations without repetition, which I show below with the formula [2]:
P(n,r) = nPr = n!/(n-r)! = 10!/(10-5)! = 10!/5! = 10*9*8*7*6 = 30240,
Where: n = the number of unique items.
For instance, n = 10 for the book example because there are 10 books. r = the size of the permutation. For example, r = 5 for the five books we want to place on the shelf. This equation works both for the complete and partial sets of permutations without repetitions, depending on the values you enter in the equation. For complete sets, n = r. Additionally, r cannot be greater than n because there are no repetitions.
Page updated
Google Sites
Report abuse