- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,055

Problem: Show that if $P(A_i)=1$ for all $i \ge 1$ then $P(\bigcap_{i=1}^{\infty}A_i)=1$.

[HR][/HR]

What is strange about this question is the first part, $P(A_i)=1$ for all $i \ge 1$. If I'm understanding this correctly that's saying that $P(A_1)=1$, $P(A_2)=1$...$P(A_n)=1$, for $n \ge i \ge 1$. This is only true if $A_1=A_2=...A_n$ because the sum of their probabilities (keeping in mind inclusion-exlusion of course) cannot be larger than 1. It's obvious that if this is the case that the intersections of all of them is 1 as well, but that's not the part that troubles me.

So is this a typo or am I misunderstanding the problem do you think?

[HR][/HR]

What is strange about this question is the first part, $P(A_i)=1$ for all $i \ge 1$. If I'm understanding this correctly that's saying that $P(A_1)=1$, $P(A_2)=1$...$P(A_n)=1$, for $n \ge i \ge 1$. This is only true if $A_1=A_2=...A_n$ because the sum of their probabilities (keeping in mind inclusion-exlusion of course) cannot be larger than 1. It's obvious that if this is the case that the intersections of all of them is 1 as well, but that's not the part that troubles me.

So is this a typo or am I misunderstanding the problem do you think?

Last edited: