1. ok we have both confused each other i think

these are the answers I get (Can't help but doubt myself )

a) Pr(A) = 7/9
b) Pr(B) = 5/9
c) Pr(A∩B) = 5/9
d) Pr(C) = 1/2
e) Pr(A∩D) = Pr(7/9 ∩ 1/3) = Pr(7/9)Pr(1/3) = 7/27

2. sorry, for part e i get 1/3

3. Originally Posted by djmccabie
ok we have both confused each other i think
a) Pr(A) = 7/9
b) Pr(B) = 5/9
c) Pr(A∩B) = 5/9
d) Pr(C) = 1/2
e) Pr(A∩D) = Pr(7/9 ∩ 1/3) = Pr(7/9)Pr(1/3) = 7/27
$P(C)=0~\&~P(D)=\frac{1}{3}$

4. I'm not sure how P(C) could be 0?

How could it be impossible?

is this a sort of continuous variable thing where as the the number gets longer, the probability tends to zero?

5. Originally Posted by djmccabie
I'm not sure how P(C) could be 0?
How could it be impossible?
is this a sort of continuous variable thing where as the the number gets longer, the probability tends to zero?
Of course it is not impossible.
That is not what probability being zero means.
Yes it has to do with continuous random variable.

The probability that Alice picks 1/2 is zero.
But she can clearly pick 1/2.

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