Question

Suppose s=x1+x2+x3+x4s=x1+x2+x3+x4 where x1,x2,x3,x4x1,x2,x3,x4 are mutually independent and xixi has the bernoulli distribution with parameter pi=i5pi=i5 for 1≤i≤4.1≤i≤4.what is p{s=1}?p{s=1}? present your answer with four digits

Answer 1

The
`X_i` are mutually independent, so that

`P(X_1=a,X_2=b,X_3=c,X_4=d)=P(X_1=a)P(X_2=b)P(X_3=c)P(X_4=d)`

for any configuration of 0s and 1s
`a,b,c,d`.

We obtain
`S=1` when exactly one of the
`X_i` are 1, and the other three are 0. There 4 possible ways of obtaining such an outcome, and they are pairwise disjoint. So

`P(S=1)=P(X_1=1,X_2=0,X_3=0,X_4=0)+\cdots+P(X_1=0,X_2=0,X_3=0,X_4=1)`

Now,

`P(X_1=1,X_2=0,X_3=0,X_4=0)=\frac15\frac35\frac25\frac15`

`P(X_1=0,X_2=1,X_3=0,X_4=0)=\frac45\frac25\frac25\frac15`

`P(X_1=0,X_2=0,X_3=1,X_4=0)=\frac45\frac35\frac35\frac15`

`P(X_1=0,X_2=0,X_3=0,X_4=1)=\frac45\frac35\frac25\frac45`

`\implies P(S=1)=(6+16+36+96)/(625)=(154)/(625)=0.2464`