Distribution of Probability With Replacement and Tree Diagram: An Example, Statistical Note 7

Among 40 participants in a training, 18 were vegetarians and 22 were non-vegetarians. 2 participants are selected at random one after another with replacement of the name of the first selected participant. Calculate the probability distribution of vegetarians.

I have taken this example from the total number of participants by food habits of my statistical note 3 to show how to visualize and calculate the probability distribution using the probability tree. I refer to my note 6 for comparison with this and identifying the difference.

There will be two consecutive selections of two participants. The first participant is selected from among the total of 40 participants. The name of the first participants selected is returned back to the sample space or the population of 40 participants. Then, the second participant will be selected again from the same 40 participants. This process is referred to as sampling with replacement. The probability in this case is called probability with replacement or independent probability.

At the first stage, there are two possibilities of randomly selecting the first participant. The first participant could be either a vegetarian or a non-vegetarian (Diagram 1). Let V1 be a simple event that the fist selected participant is a vegetarian. The marginal probability that a randomly selected participant is a vegetarian, denoted by P(V1) is 18 divided by 40, 0.45 (blue block). Similarly, the probability that a randomly selected participant is a non-vegetarian, denoted by P(NV1) is 22 divided by 40, 0.55 (green block). It is calculated also as one minus P(V1), which is equal to 0.55.

Diagram 1: First and second steps showing marginal probabilities (with replacement of the first selected participant)

At the second stage also, there will be 40 participants with four possibilities of randomly selecting a participant.  The process will be same as that at the first stage, so that P(V2) will be same as P(V1) and P(NV2) will be same as P(NV1).

Now, let me discuss about the joint probabilities selecting both the first and second participants. Let P(V1 intersection V2) or (V1∩V2) be a joint event that both the first and second selected participants are the vegetarians. The joint probability of (V1∩V2), denoted by P(V1∩V2), is the product of P(V1) and P(V2) and that is equal to 0.45 multiplied by 0.45, equal to 0.202. Following the same process, other joint probabilities are calculated, P(V1∩NV1) equal to 0.248, P(NV1∩V2) equal to 0.248, and P(NV1∩NV2) equal to 0.303.

Table 1: Discrete probability distribution of vegetarians sampled with replacement

The probability distribution of vegetarians as per the question is discussed (Table 1). Let X be an event that takes the discrete value or the number of vegetarians in two consecutive selection of participants. X takes the value 2 for the joint probability P(V1∩V2) that vegetarians are selected both times, 1 for both joint probabilities P(V∩NV) and P(NV∩V) are same as one of two participants randomly selected at the first stage or the second stage is a vegetarian if the position of the vegetarian does not matter. Thus, their probabilities are added.  X takes the value 0 for the joint probability P(NV∩NV) that non-vegetarians are selected both times.

The probability distribution of vegetarians shows that there is 30.3 percent chance that no vegetarian (or both non-vegetarians) is selected, there is 49.6 percent chance that one of two participants selected will be a vegetarian and there is 20.2 percent chance that both participants will be vegetarians or none of them will be non-vegetarian. If the probabilities are added, there is 79.9 percent chance that upto one vegetarian will be selected. There is cent percent chance of getting two or less number of vegetarians in the draw of two participants.