this post was submitted on 22 May 2024
17 points (94.7% liked)
Daily Maths Challenges
198 readers
6 users here now
Share your cool maths problems.
Complete a challenge:
- Post your solution in comments, if it is exactly the same as OP's solution, let us know.
- Have fun.
Post a challenge:
- Doesn't have to be original, as long as it is not a duplicate.
- Challenges not riddles, if the post is longer than 3 paragraphs, reconsider yourself.
- Optionally include solution in comments, let it be clear this is not a homework help forums.
- Tag [unsolved] if you don't have a solution yet.
- Please include images, if your question includes complex symbols, attach a render of the maths.
Feel free to contribute to a series by DMing the OP, or start your own challenge series.
founded 7 months ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
solution
Say Omega = N\{0}, sigma algebra is power set of N and the probability mass function is p(n)=2^-n .Then A is all the even numbers, B all numbers at least 4, C all numbers at least 5.
P(A)=sum{2^-n^ | n is even} = sum{2^-2n^ | n is at least 1} = sum{4^-n^ | n is at least 1} = 1/(1-1/4)-1=1/3
P(B) = P(N\{1,2,3}) = 1 - 1/2 - 1/4 - 1/8 = 1/8
P(C) = 1/16 similarly
P(A and B) = P(A\{2}) = 1/3 - 1/4 = 1/12 =/= P(A)P(B) therefore not independent
P(A and C) = P(A\{2,4}) = P(A)P(C) with a similar calculation and therefore independent