Reinfrank2011 wrote:
You can also just find the probability of selecting the desired individuals and then multiply that by the number of ways those selections can be ordered:
P(Andrew)*P(Not Karen)*P(Not Karen)*P(Not Karen)*(ways to order selections) = \((\frac{1}{8})*(\frac{6}{7})*(\frac{5}{6})*(\frac{4}{5})*(\frac{4!}{3!}) = (\frac{1}{14})*4 = \frac{2}{7}\)
The above is just the sum of the individual probabilities of each possible 4-person team:
Andrew, Not Karen, Not Karen, Not Karen -->
\((\frac{1}{8})(\frac{6}{7})(\frac{5}{6})(\frac{4}{5}) = \frac{1}{14}\)
Not Karen or Andrew, Andrew, Not Karen, Not Karen -->
\((\frac{6}{8})(\frac{1}{7})(\frac{5}{6})(\frac{4}{5})= \frac{1}{14}\)
Not Karen or Andrew, Not Karen or Andrew, Andrew, Not Karen -->
\((\frac{6}{8})(\frac{5}{7})(\frac{1}{6})(\frac{4}{5})= \frac{1}{14}\)
Not Karen or Andrew, Not Karen or Andrew, Not Karen or Andrew, Andrew -->
\((\frac{6}{8})(\frac{5}{7})(\frac{4}{6})(\frac{1}{5})= \frac{1}{14}\)
Thus, \((\frac{1}{14})*4=\frac{2}{7}\)
Hi
Bunuel,
Can you explain why it's 4!/3! and not just 4!. I do not understand why the "not karens" who are actually individuals and not identical items do not count determining the number of orders. Also, in case there is a page with theory about this aspect, determining the number of orders in a selection, would be great.
Thank you!