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 in which the order does not matter whether the vegetarians are sampled or not in the first or second draw.
Counting the total and favorable numbers
of outcomes constituting the specified number of objects sampled with replacement
from the finite discrete binary 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. Formula
is another means of counting the outcomes and calculating the discrete
probability distribution. I have taken this example from statistical notes 3 and
7 to show how to visualize and calculate the discrete probability distribution with
replacement using the tree diagram and formula.
Tree Diagram
There will be two consecutive
selections of two participants with replacement, referred to as sampling with
replacement. The probability each outcome is presented in Diagram 1. For
detailed discussion on how they are derived please refer to my statistical note
7.
Diagram 1: First and
second steps showing marginal probabilities (with replacement
of the first selected participant)
There are two possibilities,
either the selected participant is a vegetarian or a non-vegetarian. The joint probability
that both the first and second selected participants are the vegetarians, denoted
by P(V1∩V2), is the product of P(V1) and P(V2) which is 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 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 up to one vegetarian will be selected. There is cent percent chance of
getting two or less number of vegetarians in the draw of two participants.
Formula
This example has three
characteristic features. First, the example has a finite population of 40
participants, denoted by ‘n’. Second, each participant can be characterized as
success or failure. Since the question asks the probability of vegetarians, the
selection of a vegetarian is considered as a success, say denoted by ‘p’, and
there are 18 successes in the population; and the selection of a non-vegetarian
is considered as a failure, denoted by ‘q’. Third, each observation is
independent so that the probability of ‘p’ or ‘q’ is same for each outcome. Fourth,
a sample of two participants, denoted by ‘x’, is selected with replacement in a
way that each sample of 2 participants is equally likely to be selected.
Let X be a random variable
of interest that takes one of 0, 1 or 2 values as the number of vegetarians in
the sample of two participants sampled with replacement, denoted by ‘x’. The
probability distribution of X depends on the parameters, ‘n’ and ‘p’, and is
given by the expression
P(X=x) = C(n,x)pxqn-x
This distribution is referred to
as Binomial distribution.
In this example, n=2 and p=18/40
and ‘x’ takes the value 0 to 2. Putting these values in the above formula, one
gets
P(X=0) = [C(2,0) X (18/40)0
X (22/40)2] = (22 X 22)/ (40 X 40)= 0.303
P(X=1) = [C(2,1)
X (18/40)1 X (22/40)1] = (18 X 22)/ (40 X 40) = 0.496
P(X=2) = [C(2,2) X (18/40)2
X (22/40)0] = (18 X 18)/ (40 X 40) = 0.202
These values are equal to the ones presented in table 1 above. It
indicates that both tree diagram and formula produce the same values are useful
to calculate the discrete probability distribution with replacement.
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