Asked by jerson

Find the probability of x=4 sucesses in n=9 trials for the probability of succes p=0.6 on each trial. Round to the nearest thousanth.
Answered by Reiny
If I understand your question correctly that would be
C(9,4)(.6)^4(.4)^5 = .167
Answered by jerson
umm whats the (9,4)?
Answered by Reiny
it is C(9,4) or "9choose4"
= 9!/(4!5!)

If you are studying this level of probability, then you must be familiar with that concept.
Answered by jerson
oh yes..
but how do u plug that into a calculator?
Answered by Reiny
depends on the calculator.
On mine, which is a Casio, it is shown as
<sub>n</sub>C<sub>r</sub>

On many calculators it is paired with
<sub>n</sub>P<sub>r</sub> which is defined
as n!/(n-r)! and is used in permutations.

If you cannot find it on your calculator, look for the factorial key ! , which is found on most scientific calculators, and just perform the calculation according to the definition of C(n,r) = n!/(r!(n-r)!)
Answered by jerson
ok i figured that out.
last question where'd the (.6)^4(.4)^5
i think i know where the .6 came from but the exponent is throwing me off
Answered by Reiny
Ok, let's just consider one of the possible outcomes. S = success, F = failure

SSFFSFFFS (4 successes, 5 failures)

the prob of that specific event is
.6 x .6 x .4 x .4 x .6 x .4 x .4 x .4 x .6
= (.6)^4 x (.4)^5

but the SSFFSFFFS can be arranged in 126 ways, so .....
Answered by Reiny
Prob(success) + prob(failure) = 1
so if prob(success) = .6
then prob(failure) = 1 - .6 = .4
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