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seen May 19 at 7:47

Mar
23
comment Combinatorics Question, drawing from a pot
I read the question again and it seems d) is impossible, if we interpret it like c). So yes, you are right.
Mar
23
comment Combinatorics Question, drawing from a pot
I've edited the answer. d) and e) are the same. It is possible to use combinatorics, but you'll have to deal with conditional probabilities and probably end up with some sum, but you will not get a nice formula.
Mar
23
revised Combinatorics Question, drawing from a pot
added 137 characters in body
Mar
23
answered Combinatorics Question, drawing from a pot
Mar
22
comment Find the radius in dependency of a (the side length)
The new picture is still not enough to correct your problem. Can you tell what exactly your professor told you?
Mar
22
comment Find the radius in dependency of a (the side length)
When I said not dependent I meant $r$ cannot be expressed in terms of $a$. You can make $a$ as big as you want without affecting $r$.
Mar
22
awarded  Commentator
Mar
22
comment Find the radius in dependency of a (the side length)
Looking at picture, $r$ is not dependent of $a$.
Mar
20
revised Probability triangle question?
added 189 characters in body
Mar
20
answered Probability triangle question?
Mar
20
awarded  Citizen Patrol
Mar
20
comment Finding the probability that one of the given independent events happens
As I understood, the question didn't say "one and only one of events should happen".
Mar
20
comment Chain rule and conditional probability
What do you mean by "chain rule for conditional probability"?
Mar
20
comment Finding the probability that one of the given independent events happens
Are those events intersecting?
Mar
20
comment Finding a second linearly independent solution to a differential equation
I think you have a typo there, because $x(t)=t$ is not a solution.
Mar
20
answered Weird factoring out of a derivative
Mar
19
comment Functions satisfying $f(m+f(n)) = f(m) + n$
@Biswanath, see the answer below, it explains why you can conclude f(0)=0.
Mar
19
comment Functions satisfying $f(m+f(n)) = f(m) + n$
Taking $n=0$ gives us $f(m)=f(m+f(0))$ or $f(0)=0$. Next, taking $m=0$ we'll have $f(f(n))=n$
Mar
18
awarded  Supporter
Mar
18
awarded  Analytical