Daily Maths Challenges

183 readers

1 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 4 months ago

MODERATORS

Coin-flipping game (lemmy.world)

submitted 4 months ago* (last edited 4 months ago) by [email protected] to c/[email protected]

12 comments fedilink hide all child comments

We're playing a game. I flip a coin. If it lands on Tails, I flip it again. If it lands on Heads, the game ends.

You win if the game ends on an even turn, and lose otherwise.

Define the following events:

A: You win the game

B: The game goes on for at least 4 turns

C: The game goes on for at least 5 turns

What are P(A), P(B), and P(C)? Are A and B independent? How about A and C?

you are viewing a single comment's thread
view the rest of the comments

[–] [email protected] 2 points 2 months ago* (last edited 2 months ago)

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