SOLUTION: A committee of four people is to be selected from six Democrats and four Republicans. What is the probability that all are Democrats?
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Question 426302: A committee of four people is to be selected from six Democrats and four Republicans. What is the probability that all are Democrats?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
If you consider chosing a Democrat for any one of the positions as a success (nothing to do with politics, it is simply a way of looking at the problem), then you want the probability of 4 successes in 4 trials where the probability of success on any given trial is 0.6. I say 0.6 because there are 10 people to choose from and 6 are considered a success.
Alternatively, if your conservative politics simply won't allow you to consider selecting a Democrat as a success, you could calculate the probability of 4 failures (failure to pick a Republican) in 4 trials where the probability of success is 0.4. You will find that it reduces to the exact same calculation.
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
So, for this problem you could either compute:
\ =\ \left(4\cr 4\right\)\left(0.6\right)^4\left(0.4\right)^{0})
or
\ =\ \left(4\cr 0\right\)\left(0.4\right)^0\left(0.6\right)^{4})
And remembering that \ =\ \left(n\cr 0\right\)\ =\ 1)
and 
,
So either way, it reduces to
\ =\ P_4(0,0.4)\ =\ \left(0.6\right)^{4})
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
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