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,

$P(X_1=1,X_2=0,X_3=0,X_4=0)=\dfrac15\dfrac35\dfrac25\dfrac15$

$P(X_1=0,X_2=1,X_3=0,X_4=0)=\dfrac45\dfrac25\dfrac25\dfrac15$

$P(X_1=0,X_2=0,X_3=1,X_4=0)=\dfrac45\dfrac35\dfrac35\dfrac15$

$P(X_1=0,X_2=0,X_3=0,X_4=1)=\dfrac45\dfrac35\dfrac25\dfrac45$

$\implies P(S=1)=\dfrac{6+16+36+96}{625}=\dfrac{154}{625}=0.2464$