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.
If I understand your question correctly that would be
C(9,4)(.6)^4(.4)^5 = .167
umm whats the (9,4)?
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.
oh yes..
but how do u plug that into a calculator?
depends on the calculator.
On mine, which is a Casio, it is shown as
nCr
On many calculators it is paired with
nPr 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)!)
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
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 .....
Prob(success) + prob(failure) = 1
so if prob(success) = .6
then prob(failure) = 1 - .6 = .4
To find the probability of x = 4 successes in n = 9 trials for a probability of success p = 0.6 on each trial, we can use the binomial probability formula. The formula is:
P(x) = (nCx) * (p^x) * ((1 - p)^(n - x))
Where:
- P(x) is the probability of getting x successes
- nCx is the binomial coefficient, which represents the number of ways to choose x successes from n trials
- p^x is the probability of getting x successes
- (1 - p)^(n - x) is the probability of getting (n - x) failures
Let's calculate the probability step by step:
1. Calculate the binomial coefficient (nCx):
nCx = n! / (x! * (n - x)!)
In our case, n = 9 and x = 4, so:
9C4 = 9! / (4! * (9 - 4)!)
Simplifying:
9C4 = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1)
= 126
2. Calculate the probability of getting 4 successes (p^x):
p^x = 0.6^4 = 0.1296
3. Calculate the probability of getting 5 failures ((1 - p)^(n - x)):
(1 - p)^(n - x) = (1 - 0.6)^(9 - 4)
= 0.4^5
= 0.01024
4. Multiply all the calculated values together:
P(x) = (9C4) * (0.6^4) * (0.4^5)
= 126 * 0.1296 * 0.01024
≈ 0.165
Therefore, the probability of getting x = 4 successes in n = 9 trials with a probability of success p = 0.6 is approximately 0.165 (rounded to the nearest thousandth).