Answer:
[tex]Probability = \frac{1}{20}[/tex]
Step-by-step explanation:
Given
[tex]People = 6[/tex]
[tex]Selection = 3[/tex]
Required
Determine the probability of selecting the oldest 3
First, we need to determine the total possible selection.
Since it's a selection, we make use of combination as follows:
[tex]^nC_r = \frac{n!}{(n-r)!r!}[/tex]
In this case:
[tex]n = 6[/tex]
[tex]r = 3[/tex]
So:
[tex]^nC_r = \frac{n!}{(n-r)!r!}[/tex]
[tex]^6C_3 = \frac{6!}{(6-3)!3!}[/tex]
[tex]^6C_3 = \frac{6!}{3!3!}[/tex]
[tex]^6C_3 = \frac{6*5*4*3!}{3!3*2*1}[/tex]
[tex]^6C_3 = \frac{6*5*4}{3*2*1}[/tex]
[tex]^6C_3 = \frac{5*4}{1}[/tex]
[tex]^6C_3 = 20[/tex]
In the combination, there can be only 1 set of oldest people.
So:
[tex]Probability = \frac{1}{20}[/tex]
