How many samples of size ‘r’ can be drawn
from the population of size ‘n’?
All may know that a finite population can have many
samples of equal size, but how many?The number of samples than can be drawn from a finite population varies
whether the sampling units are drawn without or with replacement. I discussed
it in my previous Statistical Note 15 (Counting of Outcomes) using tree diagram
and formula to different examples. In this note, I take only two examples six
and eight discussing on using tree diagram and formula to draw some balls
from the population of balls without or with replacement in which the order of the
ball does not matter. Most often the position of the sampling units in a sample
does not matter. The number of samples will be important to discuss on sampling
distribution in future statistical notes.
Sampling Without Replacement
A bag contains four balls
of different colors (A of Red color, B Blue, C Green and D yellow). Select two
balls randomly one at a time and do not replace back to the bag after
selecting. Calculate the total number of outcomes containing two balls selected
without replacement in which the order of the balls does not matter.
Tree Diagram: This example
has six combinations of 12 outcomes each constituting two balls selected one at
a time without replacement from the total of four balls, Red (A), Blue (B),
Green (C) and Yellow (D) in which the order of balls does not matter. The
outcomes are indicated by numbers on the right most part of the Diagram 1.
Diagram 1: Outcomes in
Selecting Two Out of Four Balls Without Replacement in which the Order of Balls does not Matters
Formula: In the above
example, the combination formula n!/(n-r)!*r! is used (see Diagram 1 of Statistical Note 15). In this case, ‘n’
is equal to 4 and ‘r’ equal to 2 balls. Using these values in the formula, we
get 4!/(4-2)!*2!, which is equal to 4!/2!*2!, equal to 6. The combination formula
is denoted by C(n,r) or nCr also.
Thus, both tree diagram and
formula came up with six combinations of 12 outcomes in which the order of the
balls does not matter. Also refer to Diagram 1 and Example 6 of the Statistical
Note 15 for detailed elaboration of this case.
Sampling With Replacement
A bag contains three balls of different colors (A of Blue
color, B Green and C Red). Select two balls one randomly one at a time and
replace back to the bag after selecting. Calculate the total number of outcomes
containing two balls selected with replacement in which the order of balls does
not matter.
Tree Diagram: This example
has six combinations of nine outcomes each constituting two of all three balls,
Blue (A), Green (B) and Red (C) selected one at random with replacement in
which the order of balls does not matter. Each combination is indicated by the
ball letters in a box and a number on the right most part of the Diagram 2.
Diagram 2: Outcomes in Selecting Two Out of Three Balls With Replacement in which
the Order of Balls does not Matter
Formula: In above example,
a formula (n+r-1)!/(n-1)!*r! is used (see Diagram 1 of Statistical Note 15). In this case, ‘n’
is equal to 3 and ‘r’ is equal to 2 balls. Using these values in the formula,
we get 4!/2!*2!, which is equal to 6.
Thus, both tree diagram and formula came up with six
combinations of nine outcomes consisting two balls in which the order of balls
does not matter. Also refer to Example 8 of the Statistical Note 15 for
detailed elaboration of this case.
The examples showed that
the number of samples without or with replacement differed in which the
position of the sampling units did not matter.
