Counting of Outcomes Using Tree Diagram and Formula, Statistical Note 15

Counting the total and favorable numbers of outcomes constituting the objects selected is important to calculate the probability of favorable events. An appropriate process of counting needs to be followed to get an appropriate result. Tree diagram is an important means to visualize and count the number of outcomes. However, as the number of objects available and selected increases, the diagram becomes complicated and it will be less possible to show all outcomes in the diagram. Thus, in that case one needs to use the formula to count the number of outcomes. This article presents both tree diagram and formula to exemplify the appropriate process of counting.

1. Conditions for Counting of Outcomes

Population or Sample: Population is the entire count of all objects or the sampling units, also called the sample space. Population census is an example in which all people are counted. Not feasible to study the entire population. Usually, a sample constituting some sampling units or objects is drawn from the population. Household survey is a common example of selecting some households from the entire list of households. This article considers that all objects are different, that is both population or sample constitute objects that are different from one another.

Without or With Replacement: Whether all or some objects are selected one may be interested to know whether the objects are drawn without or with replacement. In selecting without replacement, the object is not replaced back once selected. Unlike, the object is replaced back after it is selected so that the object is available for following selections. These processes are also referred to as sampling without or with replacement.

Order or Non-order: Whether an object is selected without or with replacement, if the order of objects matters, the club of objects forms a separate outcome. This is also called Ordered Sampling. This leads to the calculation of Permutation. If the order of objects does not matter, the outcomes constituting the objects are same whether the objects occur in one order or another. This is called Non-ordered or Unordered Sampling.

2. Formula

Overall, the multiplication principle is applied, which is the product of the number of objects available for selection every time to the times the number of objects need to be selected to get the total number of outcomes whether all objects (population) or some (sample) of them need to be selected. The formula differs from one condition to another as presented in Diagram 1.

Diagram 1: Formula of Counting the Number of Outcomes by Sampling technique and Order of Objects

Let ‘n’ be the total number of objects (population) and ‘r’ be the size of the sample drawn. If out of ‘n’ objects ‘r’ objects (sample) are selected at random without replacement such that the order or the order of each object matters is given by the Permutation formula, denoted by ‘ⁿP_r’or P(n,r) is given by n!/(n-r)!, that is ‘n’ Factorial divided by ‘n-r’ Factorial. If the order does not matter, the total number of outcomes constituting the objects selected without replacement is counted using the Combination formula, denoted by ‘ⁿC_r’ C(n,r), (n_r) is given by n!/ (n-r)! r!, that is ‘n’ Factorial divided by ‘(n-r)’ Factorial divided by ‘r’ Factorial. If out of ‘n’ objects ‘r’ objects are selected at random with replacement such that the order of each object matters is given by the Exponential formula, denoted by ‘n^r’or ‘n power r’. If the order does not matter, the total number of outcomes is counted using the formula, denoted by ‘^n+r-1C_r’ C(n+r-1,r), (n+r-1 _r) is given by (n+r-1)!/(n-1)! r!, that is ‘n+r-1’ Factorial divided by ‘(n-1)’ Factorial divided by ‘r’ Factorial.

3. Example

The examples use tree diagram and formula to visualize and count the number of outcomes using all objects or some of them selected randomly without or with replacement considering or not the order of objects in the outcomes.

Example 1:         Outcomes of All Objects Selected Without Replacement in which the Order of Objects Matters

A bag contains four balls of different colors (A of Red color, B Blue, C Green and D yellow). Select the first ball at random and do not replace back to the bag. Select remaining balls one at a time following the same process. Calculate the total number of possible outcomes containing all four balls selected without replacement in which the order of the balls matters.

Tree Diagram: In this example that I discussed in my statistical note 8, there are 24 unique outcomes each constituting all four balls, Red (A), Blue (B), Green (C) and Yellow (D) selected one at a time without replacement and arranged in which the order of balls matters. The outcomes are indicated by numbers on the right most side of Diagram 2.

Diagram 2: Outcomes of All Objects Selected Without Replacement in Which the Order of Objects Matters

Formula: In above example, the Permutation Formula n!/(n-r)! is used (see formula listed in the cell between ‘Without Replacement’ and ‘Matters order of objects’ in Diagram 1). In this case, all ‘n’ objects are selected so that both ‘n’ and ‘r’ are equal to 4 balls. Using the formula, we get 4!/0!, which is equal to 4! as 0! is equal to 1. On further calculation one gets 4 X 3 X 2 X 1 equal to 24, applying the multiplication principle. Thus, both tree diagram and formula came up with 24 outcomes in which the order of each ball matters.

Example 2:         Outcomes of All Objects Selected Without Replacement in which the Order of Objects does not Matter

A bag contains four balls of different colors (A of Red color, B Blue, C Green and D yellow). Select the first ball at random and do not replace back to the bag. Select remaining balls one at a time following the same process. Calculate the total number of outcomes containing all four balls selected without replacement in which order of the balls does not matter.

Tree Diagram: In this example that I discussed in my statistical note 10, there is only one combination of all 24 outcomes each constituting all four balls, Red (A), Blue (B), Green (C) and Yellow (D) selected one at random without replacement and the order of balls does not matter. Because every of 24 outcomes has all four balls arranged in one way or the other. This is indicated by 1 number on the right most part of the diagram 1 to mean one combination in Diagram 3.

Diagram 3: Combination of All Objects Selected Without Replacement in which the Order of Objects does not Matter

Formula: In above example, the Combination Formula n!/(n-r)!*r! is used  (see formula listed in the cell between ‘Without Replacement’ and ‘Does Not Matters Order of Objects’ in Diagram  1). In this case, all ‘n’ objects are selected so that both ‘n’ and ‘r’ are equal to 4 balls. Using this value in the formula, we get 4!/0!*4!, which is equal to 1 as 0! Is equal to 1. Thus, both tree diagram and formula came up with one combination of same 24 outcomes in which the order of balls does not matter.

Example 3: Outcomes of All Objects Selected With Replacement in which the Order of Objects Matters

A bag contains three balls of different colors (A of Blue color, B Green and C Red). Select all three balls randomly one at a time and replace back to the bag after selecting. Calculate the total number of combination of possible outcomes containing all three balls selected with replacement in which the order of balls matters.

Tree Diagram: In this example that I discussed in my statistical notes 10, there are 27 outcomes each constituting all three balls, Blue (A), Green (B) and Red (C) selected one at random with replacement and the order of balls matters. Because every of nine outcomes has all three balls arranged in one way or the other. The unique outcomes are indicated by numbers on the right most part of the Diagram 4.

Diagram 4: Outcomes of All Objects Selected With Replacement in which Order of the Objects Matters

Formula: In above example, the Exponential Formula n^r is used (see formula listed in the cell between ‘With Replacement’ and ‘Matters Order of Objects’ in Diagram  1). In this case, all ‘n’ objects are selected so that both ‘n’ and ‘r’ are equal to 3 balls. Using this value in the formula, we get 3³, which is equal to 27. Thus, both tree diagram and formula came up with 27 unique outcomes of balls in which the order of balls matters.

Example 4: Outcomes of All Objects Selected With Replacement in which the Order of Objects does not Matter

A bag contains three balls of different colors (A of Blue color, B Green and C Red). Select all three balls randomly one at a time and replace back to the bag after selecting. Calculate the total number of combination of possible outcomes containing all three balls selected with replacement in which the order of balls does not matter.

Tree Diagram: In this example that I discussed in my statistical notes 10, there are 10 combinations of all 27 outcomes each constituting all three balls, Blue (A), Green (B) and Red (C) selected one at random with replacement in which the order of balls does not matter. Each combination is indicated by a number on the right most part of the Diagram 5.

Diagram 5: Outcomes of All Objects Selected With Replacement in which the Order of Objects does not Matter

Formula: In above example, a formula (n+r-1)!/(n-1)!*r! is used (see formula listed in the cell between ‘With Replacement’ and ‘Does Not Matters Order of Objects’ in Diagram  1). In this case, both ‘n’ and ‘r’ are equal to 3 balls. Using these values in the formula, we get 5!/2!*3!, which is equal to 10. Thus, both tree diagram and formula came up with 10 combinations of 27 outcomes consisting of three balls in which the order of balls does not matter.

Example 5: Some Objects (Sample) Selected Without Replacement from All Objects (Population) in which Order of Objects Matters

A bag contains four balls of different colors (A of Red color, B Blue, C Green and D yellow). Select two balls randomly one at a time and do not replace back to the bag after selecting. Calculate the total number of combination of possible outcomes containing two balls selected without replacement in which the order of the balls matters.

Tree Diagram: In this example that I discussed in my statistical notes 10 and 11, the process is same as that discussed above in Example 1. There are 12 unique outcomes each constituting two balls selected one at a time without replacement from the total of four balls, Red (A), Blue (B), Green (C) and Yellow (D) in which the order of balls matters. The outcomes are indicated by numbers on the right most part of the Diagram 6.

Diagram 6: Outcomes in Selecting Two Out of Four Balls Without Replacement in which the Order of Balls Matters

Formula: In above example, as in Example 1 the Permutation Formula n!/(n-r)! is used (see formula listed in the cell between ‘Without Replacement’ and ‘Matters order of objects’ in Diagram  1). In this case, ‘n’ is equal to 4 balls and ‘r’ is equal to 2 balls. Using these values to the formula, we get 4!/2!, which is equal to 12. Thus, both tree diagram and formula came up with 12 unique outcomes in which the order of each ball matters.

Example 6: Some Objects (Sample) Selected Without Replacement from All Objects (Population) in which Order of Objects does not Matter

A bag contains four balls of different colors (A of Red color, B Blue, C Green and D yellow). Select two balls randomly one at a time and do not replace back to the bag after selecting. Calculate the total number of outcomes containing two balls selected without replacement in which the order of the balls does not matter.

Tree Diagram: In this example that I discussed in my statistical notes 10 and 11 and in Example 2 above, there are six combinations of 12 outcomes each constituting two balls selected one at a time without replacement from the total of four balls, Red (A), Blue (B), Green (C) and Yellow (D) in which the order of balls does not matter. The outcomes are indicated by numbers on the right most part of the Diagram 7.

Diagram 7: Outcomes in Selecting Two Out of Four Balls Without Replacement in which the Order of Balls Matters

Formula: In the above example, the combination formula n!/(n-r)!*r! is used (see formula listed in the cell between ‘Without Replacement’ and ‘Does Not Matters Order of Objects’ in Diagram  1). In this case, ‘n’ is equal to 4 and ‘r’ equal to 2 balls. Using these values in the formula, we get 4!/(4-2)!*2!, which is equal to 4!/2!*2!, equal to 6. Thus, both tree diagram and formula came up with six combinations of 12 outcomes in which the order of the balls does not matter.

Example 7: Some Objects (Sample) Selected With Replacement from All Objects (Population) in which the Order of Objects Matters

A bag contains three balls of different colors (A of Blue color, B Green and C Red). Select two balls one randomly one at a time and replace back to the bag after selecting. Calculate the total number of outcomes containing two balls selected with replacement in which the order of balls matters.

Tree Diagram: In this example that I discussed in my statistical notes 10 and 12 and Example 3 above, there are nine outcomes each constituting all three balls, Blue (A), Green (B) and Red (C) selected one at random with replacement and the order of balls matters.. The unique outcomes are indicated by numbers on the right most part of the Diagram 8.

Diagram 8: Outcomes in Selecting Two Out of Three Balls With Replacement in which the Order of Balls Matters

Formula: In the above example, the Exponential Formula n^r is used (see formula listed in the cell between ‘With Replacement’ and ‘Matters Order of Objects’ in Diagram  1). In this case, both ‘n’ is equal to 3 and ‘r’ equal to 2 balls. Using these values in the formula, we get we get 3², which is equal to 9. Thus, both tree diagram and formula came up with nine unique outcomes of balls in which the order of balls matters.

Example 8: Some Objects (Sample) Selected With Replacement from All Objects (Population) in which the Order of Objects does not Matter

A bag contains three balls of different colors (A of Blue color, B Green and C Red). Select two balls one randomly one at a time and replace back to the bag after selecting. Calculate the total number of outcomes containing two balls selected with replacement in which the order of balls does not matter.

Tree Diagram: In this example that I discussed in my statistical notes 10 and 12 and Example 4 above, there are six combinations of nine outcomes each constituting all three balls, Blue (A), Green (B) and Red (C) selected one at random with replacement and the order of balls does not matters. Each combination is indicated by a number on the right most part of the Diagram 9.

Diagram 9: Outcomes in Selecting Two Out of Three Balls With Replacement in which the Order of Balls does not Matter

Formula: In above example, a formula (n+r-1)!/(n-1)!*r! is used (see formula listed in the cell between ‘With Replacement’ and ‘Does Not Matters Order of Objects’ in Diagram  1). In this case, ‘n’ is equal to 3 and ‘r’ is equal to 2 balls. Using these values in the formula, we get 4!/2!*2!, which is equal to 6. Thus, both tree diagram and formula came up with six combinations of 12 outcomes consisting two balls in which the order of balls does not matter.