Answer:
a. 0.689
b. 0.8
c. 0.427
Step-by-step explanation:
The given scenario indicates hyper-geometric experiment because because successive trials are dependent and probability of success changes on each trial.
The probability mass function for hyper-geometric distribution is
P(X=x)=kCx(N-k)C(n-x)/NCn
where N=4+3+3=10
n=2
k=4
a.
P(X>0)=1-P(X=0)
The probability mass function for hyper-geometric distribution is
P(X=x)=kCx(N-k)C(n-x)/NCn
P(X=0)=4C0(6C2)/10C2=15/45=0.311
P(X>0)=1-P(X=0)=1-0.311=0.689
P(X>0)=0.689
b.
The mean of hyper-geometric distribution is
μx=nk/N
μx=2*4/10=8/10=0.8
c.
The variance of hyper-geometric distribution is
σx²=nk(N-k).(N-n)/N²(N-1)
σx²=2*4(10-4).(10-2)/10²*9
σx²=8*6*8/900=384/900=0.427