Multi-Category Discrete Probability Distribution of Sampling With Replacement and Tree Diagram and Formula, Statistical Note 19

Roll an unbiased die twice, note which face turns up each time a die is rolled. Calculate the probability distribution of a face with number one.

Counting the total and favorable numbers of outcomes constituting the specified number of objects sampled with replacement from the finite population is important to calculate the probability of favorable events. Tree diagram is an important means to visualize, count the number of outcomes and calculate the probability. As the number of times the objects are selected increases, the diagram becomes complicated and it will be less possible to show all outcomes in the diagram. Thus, Formula is used for counting the outcomes and calculating the discrete probability distribution. I discuss how to visualize and calculate the multi-category discrete probability distribution with replacement using the tree diagram and formula.

Tree Diagram

A dice has six faces with nominal number one (A), two (B), three (C), four (D), five (E) and six (F). Every roll is independent and in every roll any of all mutually exclusive and possible values are likely to occur. The probability of a face in the first roll is one divided by six. Two independent rolls of a dice, also known as the sampling with replacement, have altogether thirty six outcomes (Diagram 1).

Diagram 1: First and second steps showing marginal probabilities (sampling with replacement) in two rolls of a dice

Total of 36 outcomes are grouped as shown in Table 1. One outcome has both faces with nominal number one (A²), yellow cell. Five outcomes have same faces with other nominal numbers (B², C², D², E², and F²), blue cells. 10 outcomes have one face with nominal number one and another face with other nominal numbers, green cells. 20 outcomes have both faces with other two nominal numbers than one in two rolls of a dice, pink cells.

Table 1: Outcomes of two rolls of a dice (sampling with replacement)

The joint probability that both the first and second rolls have the same faces with nominal number one, denoted by P(A1∩A2), is the product of P(A1) and P(A2) which is equal to one divided by 36 or 0.027. Following the same process, other joint probabilities of the same values are calculated.

Let X be a random variable that takes the value 0, 1 or 2 as the number of faces with nominal number one in two rolls of a dice.

As discussed above, 25 outcomes have both faces with other nominal numbers than one in both rolls of a dice, each outcome with the joint probability of one divided by 36. Thus, P(X=0) will be 25 divided by 36, equal to 0.694 (Table 2).

10 outcomes have one face with nominal number one and another face with other nominal numbers, each outcome with the joint probability of one divided by 36. Thus, P(X=1) will be 10 divided by 36, equal to 0.278 (Table 2).

One outcome has both faces with nominal number one, with the joint probability of one divided by 36. Thus, P(X=2) will be one divided by 36, equal to 0.028 (Table 2).

Table 2: Discrete probability distribution of a face with nominal number one in two rolls of a dice sampled with replacement

The probability distribution of faces with nominal number one shows that there is 69.4 percent chance that face with other nominal numbers are sampled in two rolls of a dice, there is 97.8 percent chance that up to one  face with nominal number one will be selected. There is cent percent chance of getting two or less number of faces with nominal number one in two rolls of a dice.

Formula

This example has five characteristic features. First, the example has a finite population, denoted by ‘n’. Second, there are more than two possible categories that are mutually exclusive in each trial or experiment. Third, each trail is independent because all categories could have same or different chance of occurrence. Fourth, each category can have equal or different probability of occurrence. Fifth, each category can occur ‘0’ to ‘n’ times in ‘n’ independent trials.

Let X be a random variable of interest that takes the value 0 to n, denoted by ‘x’ in the sampling with replacement. Let ‘c’ denote the number of outcome categories that the 1^st category occur n₁ times to c^th category occurs n_c times with their probabilities denoted by p₁, p₁ to p_c.

The probability distribution of X is given by the expression

P(X=x) = P(n₁,n₂,...,n_c)= [n!/(n₁! n₂! n₃! .. n_c!)](p₁ⁿ₁ p₂ⁿ₂ p₃ⁿ₃ …p_cⁿ_c)                    

This distribution is referred to as Multinomial distribution. The term [n!/(n₁! n₂! n₃! .. n_c!)] is referred to as the Multinomial coefficient.

In this example, n=2 and each of faces A to F occurs 0 to 2 times (no face with the nominal number of interest to 2 faces with that nominal number), each with the probabilities p1=p1= p2= p3= p4= p5= p6=1/6. This example has four groups of outcomes as indicated by cells of different colors in Table 1.

Group One Outcomes:

Five outcomes do not have the face with nominal number one (A) but have faces with other nominal numbers in both rolls of a dice : both times face with number two (B²) and others namely C², D², E² and F². For each outcome, the probability that both rolls have others than the face with number one (A) is calculated using the formula. Taking the outcome B² (face with nominal number two in both rolls of a dice) in the above formula, one gets

P(0,2,0,0,0,0)= [2!/(0!2! 0! 0! 0! 0!)][(1/6)⁰*(1/6)²*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰] = 1/36

Thus, the total probability that both rolls do not have the face with number one is five times of P(0,2,0,0,0,0), equal to 5/36.

Group Two Outcomes:

10 outcome combinations do not have the face with nominal number one (A) but have faces with different nominal numbers in both rolls of a die: BC, BD, BE, BF, CD, CE, CF, DE and DF. For each outcome, the probability that both rolls have others than the face with number one (A) is calculated, taking the outcome BC, as:

P(0,1,1,0,0,0)= [2!/(0!1! 1! 0! 0! 0!)][(1/6)⁰*(1/6)¹*(1/6)¹*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰] = 2/36

Thus, the total probability that both rolls do not have the face with number one is ten times of P(0,1,1,0,0,0), equal to 20/36.

Group Three Outcomes:

Five outcome combinations have one face with number one in any of two rolls of a die: AB, AC, AD, AE, and AF. For each outcome, the probability that one of two rolls has the face with number one (A) is calculated as:

P(1,1,0,0,0,0)= [2!/(1!1! 0! 0! 0! 0!)][(1/6)¹*(1/6)¹*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰] = 2/36

Thus, the total probability that one of two rolls has the face with number one (A) is five times of P(0,1,1,0,0,0), equal to 10/36.

Group Four Outcome:

Only one outcome has the face with number one (A) in both rolls of a die. For that outcome, the probability is calculated as:

P(2,0,0,0,0,0)= [2!/(2!0! 0! 0! 0! 0!)][(1/6)²*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰*(1/6)⁰] = 1/36

The probability of group one and two outcomes are added to calculate the probability that both faces are with different nominal numbers in two rolls of a dice, which is equal to the first row probability in Table 2. The probability of group three and four outcomes are equal to the second and third row probabilities in Table 2 respectively. It indicates that both tree diagram and formula produce the same values and are useful to calculate the multi-category discrete probability distribution of samples drawn with replacement.