Given:
E1,E2...En, are independent events
\(\displaystyle{P}{\left({E}{i}\right)}=\frac{{1}}{{{i}+{1}}}\)

To proof: \(\displaystyle{P}{\left({E}{1}{U}\ldots{U}{E}{n}\right)}=\frac{{n}}{{{n}+{1}}}\)

Use the Complement rule: \(\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)

PSKP((Ei)^c)=1-P(Ei) =1-(1/(i+1)) =(i+1)/(i+1)-(1/(i+1)) =i/(i+1)ZSK

Since the events E1, E2..., En are independent, the events \(\displaystyle{\left({E}{1}\right)}^{{c}},{\left({E}{2}\right)}^{{c}}\ldots{\left({E}{n}\right)}^{{c}}\) are alslo independent.

We can use the multification rule for independent events PSKP(A ∪ B)= P(A and B) = P(A)xP(B)ZSK

PSKP((E1)^c)⋂...⋂((En)^c)=P(E1)^c*...*P((En)^c) =1/(1+1)*(2/(2+1))*(3/(3+1))*...*(n/(n+1) =1/2*2/3*3/4*...*n/(n+1) =(1(2)(3)(4)...(n))/(2(3)(4)...(n+1)) =1/(n+1)ZSK

Use the complement rule: \(\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)

PSKP(E1∪...∪En)=P(((E1)^c)⋂...⋂(En)^c)) =1-P(((E1)^c)⋂...⋂(En)^c)=1-(1/(n+1))=n/(n+1)ZSK

To proof: \(\displaystyle{P}{\left({E}{1}{U}\ldots{U}{E}{n}\right)}=\frac{{n}}{{{n}+{1}}}\)

Use the Complement rule: \(\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)

PSKP((Ei)^c)=1-P(Ei) =1-(1/(i+1)) =(i+1)/(i+1)-(1/(i+1)) =i/(i+1)ZSK

Since the events E1, E2..., En are independent, the events \(\displaystyle{\left({E}{1}\right)}^{{c}},{\left({E}{2}\right)}^{{c}}\ldots{\left({E}{n}\right)}^{{c}}\) are alslo independent.

We can use the multification rule for independent events PSKP(A ∪ B)= P(A and B) = P(A)xP(B)ZSK

PSKP((E1)^c)⋂...⋂((En)^c)=P(E1)^c*...*P((En)^c) =1/(1+1)*(2/(2+1))*(3/(3+1))*...*(n/(n+1) =1/2*2/3*3/4*...*n/(n+1) =(1(2)(3)(4)...(n))/(2(3)(4)...(n+1)) =1/(n+1)ZSK

Use the complement rule: \(\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)

PSKP(E1∪...∪En)=P(((E1)^c)⋂...⋂(En)^c)) =1-P(((E1)^c)⋂...⋂(En)^c)=1-(1/(n+1))=n/(n+1)ZSK