Outcomes With Replacement and Tree Diagram, Statistical Note 9

A bag contains three balls of different colors (A of Blue color, B Green and C Red). Select the first ball at random and 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 three balls.

Calculating the total number of outcomes using all objects selected one at a time with replacement is important to calculate the probability of events. Also, important is a mathematical concept of the ‘Exponent’ used in identifying the total number of possible outcomes and calculating the probability of events. I show the total number outcomes using the tree diagram 1.

Diagram 1: Outcomes With Replacement

There will be three stages required to arrange all three balls sequentially. At the first stage or draw, there are three possibilities of randomly selecting the first ball. The first ball could be one of Blue (A), Green (B) or Green (C) balls shown by boxes of those letters and colors in Diagram 1. Once the first ball, Blue (A), is drawn and replaced back, there will be again all three balls in the bag, A, B and C. At the second stage or draw also, there are three possibilities, one of A, B or C balls. If the second ball drawn is again A and replaced back, there will be again be three balls in the bag leaving three possibilities A, B or C for the third stage or draw. If the third ball drawn is again Blue (A) the first outcome constituting all three balls is AAA which is numbered 1 on the right most side of the diagram 1. Following the same process, if the third ball drawn is Green (B), the second outcome is AAB, numbered 2 in the diagram 1. In the same way, there will be altogether 27 outcomes, each constituting three balls. Every outcome is unique in terms of position of three balls. Note that this uniqueness of an outcome because of the position of each of three balls is called Permutation. We’ll discuss more about Permutation in some other notes.

The total number of possible outcomes is calculated by multiplying three possibilities at each of the first second and third stage. This is calculated as 3 X 3 X 3 equal to 27 represented by the ‘three to the power three’ or ‘three to the third power’ or ‘three cubed’, denoted by 3³. In this formula, the first three is called the ‘Base’ and the upper three is called the ‘Exponent’ or ‘Index’ or ‘Power’. Thus, the concept of exponent is used to calculate the number of possible outcomes given the number of balls, a case in which a ball is replaced back once drawn. This could be termed as outcomes with replacement.

How many will be the outcomes with replacement if there are five balls? Can you calculate using the factorial formula? Now, you could use the formula 5⁵, which is 5 X 5 X 5 X 5 X 5 equal to 3125. Can you show in the diagram? Perhaps, not. 

Thus, the diagrammatic logic is important to understand how to calculate the number possible outcomes with replacement. As the number of balls or entities increases, the diagram becomes complicated and it will be less possible to show all outcomes in the diagram. With this ‘Exponent’ formula, one can easily calculate the number of possible outcomes given the number of balls or entities.