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Hi

I have this here probability axiom which I'm not sure what I have understood correctly.

Let [tex]B_1 \ldots B_n[/tex] be independent events

Then [tex]P(B_1 \mathrm{U} \ldots \mathrm{U} \ B_n) = 1[/tex] which is the same as

[tex]P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]

I would like to show that this only is valid if [tex]1 \leq k \leq n[/tex] such that

P(B_k) = 1.

Proof:

If [tex]1 \leq k \leq n[/tex], then [tex]P(E) = 1 [/tex](where E is the probability space).

Thereby it follows that [tex]P(B_1 U \ldots U \ B_n) = P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]

This can be written as the [tex]\sum_{n=1} ^{k} P(B_{n+1}) = P(B_{k}) =1[/tex]

Am I on the right track here?

Best Regards

Fred

I have this here probability axiom which I'm not sure what I have understood correctly.

Let [tex]B_1 \ldots B_n[/tex] be independent events

Then [tex]P(B_1 \mathrm{U} \ldots \mathrm{U} \ B_n) = 1[/tex] which is the same as

[tex]P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]

I would like to show that this only is valid if [tex]1 \leq k \leq n[/tex] such that

P(B_k) = 1.

Proof:

If [tex]1 \leq k \leq n[/tex], then [tex]P(E) = 1 [/tex](where E is the probability space).

Thereby it follows that [tex]P(B_1 U \ldots U \ B_n) = P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]

This can be written as the [tex]\sum_{n=1} ^{k} P(B_{n+1}) = P(B_{k}) =1[/tex]

Am I on the right track here?

Best Regards

Fred

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