Find C(n,x) p^xq^n-x for the given values of n,x, and p.
n=4, x=2, p=1/3
Did I set this up right?
To find q: 1-1/3=2/3
n-x=4-2=2
P(X=0) = C(4,0) (1/3)^0 (2/3)^2
Nevermind I actually plugged in the wrong number for x. Please ignore.
more standard notation
chance that it will happen this time = p = 1/3
chance that it will not = q = 1-p = 2/3
chance that it will happen r times in n trials
= C(n,r) p^r q^(n-r)
for it to happen twice in 4 trials
n = 4
r = 2
yes you would have
C(4,2) (1/3)^2 (2/3)^2
C(4,2)= 4!/[(4-2)!(2!)] = 4*3*2/[ 2*2] = 6
so
6 *(1/9)(4/9) = 2/3 * 4/9 = 8 / 27
Yes, you have set up the expression correctly.
To find q, you subtract p from 1. In this case, p is 1/3, so q would be 1 - 1/3 = 2/3.
The formula you are using is for the binomial coefficient C(n, x), also known as "n choose x". It represents the number of ways to choose x items from a set of n items without regard to their order.
The formula C(n, x) is given by:
C(n, x) = n! / (x! (n-x)!)
In your case, n = 4 and x = 2.
Plugging in these values, we get C(4, 2) = 4! / (2! (4-2)!) = 4! / (2! 2!) = (4x3x2x1) / (2x1x2x1) = 6.
So, C(4, 2) = 6.
Now, we can substitute the values of n, x, and p in the expression C(n, x) * p^x * q^(n-x).
C(n, x) = 6, p = 1/3, and q = 2/3.
Plugging in these values, we get:
C(4, 2) * (1/3)^2 * (2/3)^(4-2) = 6 * (1/3)^2 * (2/3)^2 = 6 * (1/9) * (4/9) = 24/81 = 8/27.
So, the value of C(n, x) * p^x * q^(n-x) when n = 4, x = 2, and p = 1/3 is 8/27.