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
2 answers
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
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
1 answer