Distribution of Probability Without Replacement and Tree Diagram: An Example, Statistical Note 6

Among 40 participants in a training, 18 were vegetarians and 22 were non-vegetarians. 2 participants are selected at random one after another without 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.

There will be two consecutive selections of two participants. The first participant is selected from among the total of 40 participants and then the second participant will be selected from the remaining 39 participants without putting back the first participant in the list. This process is referred to as sampling without replacement. The probability in this case is called probability without replacement or dependent 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 and conditional probabilities (without replacement of the first selected participant)

At the second stage, 39 participants are left with four possibilities of randomly selecting the second participant. The first two possibilities are discussed and remaining two possibilities will follow the same process.

Let V2/V1 be an event that the second selected participant is also a vegetarian given the first participant is a vegetarian.  Now, the conditional probability of V2/V1, denoted by P(V2/V1), is 17 vegetarians left divided by total of 39 participants left, 0.43 (grey block). Now, the second possibility is discussed. Let NV1/V1 be an event that the second selected participant is a non-vegetarian given the first participant is a vegetarian. The conditional probability of NV1/V1, denoted by P(NV1/V1), is 22 non-vegetarians divided by total of 39 participants left, 0.56 (yellow block). Following the same process, P(V1/NV1) is 0.46 (red block) and P(NV2/NV1) is 0.54 (purple block).

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/V1) and that is equal to 0.45 multiplied by 0.43, equal to 0.194. Following the same process, other joint probabilities are calculated, P(V1∩NV1) equal to 0.252, P(NV1∩V1) equal to 0.253, and P(NV1∩NV2) equal to 0.297.

  

Table 1: Discrete probability distribution of sampled vegetarians

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 the times, 1 for both joint probabilities P(V1∩NV1) and P(NV1∩V1), which are same as one of two participants are 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(NV1∩NV2) that none of two selected participants is a vegetarian.

The probability distribution of vegetarians shows that there is 29.7 percent chance that no vegetarian (or both non-vegetarians) is selected, there is 50.5 percent chance that one of two participants selected will be a vegetarian and there is 19.4 percent chance that both participants will be vegetarians or none of them will be non-vegetarian. If the probabilities are added, there is 85.2 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.