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Suppose P([1/n,1/2])<= 1/3 for all n = 1,2,3,...

a) Must we have P((0,1/2]) <= 1/3 ?

b) Must we have P([0,1/2]) <= 1/3 ?

My answer - Please correct me if i am wrong

a)

P([1/n,1/2])<= 1/3 for all n = 1,2,3,...

lim P([0,1/2])

n-> infinity

so then P((0,1/2]) is a subset of P([1/n,1/2]) for all n = 1,2,3,...

because any number in (0,1/2] can lie in [1/n,1/2] provided the correct n value is in -> 1,2,3...

so true

b)

P([1/n,1/2])<= 1/3 for all n = 1,2,3,...

lim P([0,1/2])

n-> infinity

so then P([0,1/2]) does not equal P([1/n,1/2]) for all n = 1,2,3,...

because as n approaches infinity, the limit of 1/n goes to 0, but doesn't touch 0

so false

2.

Suppose P((0,1/2]) = 1/3. Prove that there is some n such that P([1/n,1/2]) > 1/4.

lim P([1/n,1/2]) = P([0,1/2]) > 1/3 > 1/4

n -> infinity

i don't think this is entirely right, i sorta guessed...

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# Homework Help: Help with my 2 easy but hard probability questions!

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