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2m
comment Probability of Probabilities :)
$15$ matches, $2^{15}$ tickets.
5m
comment Probability of picking all white marbles?
Sorry, yes, upside down, the number of "favourables" is $\binom{5}{5}\binom{n-k}{k-5}$.
8m
reviewed Leave Open Doubt with Absolute Value Inequality
9m
reviewed Leave Open Find the smallest natural number $n$
10m
reviewed Leave Open compositions of n with k even summands and compositions of n-k with k odd summands
23m
comment Probability of picking all white marbles?
There are $\binom{n}{k}$ equally likely ways to grab $k$ marbles. There are $\binom{5}{5}\binom{k-5}{n-k}$ ways to grab $k$ marbles, among them all the white. Of course $k\ge 5$,
35m
comment Distribution of Summation of two discrete random variables
Note that the support set for the sum is $7$ to $19$.
50m
comment Dice Roll Probabilities
The problem was stated clearly, and that is exactly the problem that I solved.
52m
revised Dice Roll Probabilities
added 193 characters in body
1h
answered Dice Roll Probabilities
1h
comment Distribution of Summation of two discrete random variables
I take it that we know nothing about the distributions of $X$ and $Y$ (apart from the supports), and we do not know whether or not they are independent.
1h
comment Is there always a square between two consecutive cubes?
Well, you attempted it, in that you undoubtedly scanned mentally the first few cubes. That is a very important first step, which I should have put down, since I did that reality check before starting to write the proof.
1h
comment Is there always a square between two consecutive cubes?
The answer was clear to you. The rest was grind. Almost anything works, since after a while there are many squares between consecutive cubes. So the required inequalities were not going to give trouble.
1h
answered Is there always a square between two consecutive cubes?
2h
comment Numbers of factors of (n)(n+1)/2 is product of exponents?
@CuriousGeorge: Say we are making a factor of $3^4 \cdot 7^1\cdot 13^8$. First we decide how many $3$'s the factor will have, The factor can have $0$ $3$'s, or $1$ $3$, or $2$, or $3$, or $4$, a total of $5$ choices. For every such choice, there are $2$ ways to decide how many $7$'s the factor will have, and then $9$ ways to decide how many $13$'s, for a total of $(5)(2)(9)$.
2h
comment Probability of Winning a Toss
If the tosses are independent, no one can assure you that if you toss $Y$ or fewer times you will win. At most they can say that you will win with probability say $0.99$. If possible, the situation should be described more precisely.
2h
comment Length of a belt?
We need to know two things, the radius, and something telling us where the kink is, maybe its distance from the centre, or the angle at the kink, anyway, something.
2h
comment Numbers of factors of (n)(n+1)/2 is product of exponents?
Are you wondering about the number of (positive) divisors of $p_1^{a_1}\cdots p_k^{a_k}$ being $(a_1+1)(a_2+1)\cdots(a_k+1)$?
2h
comment Confidence interval of a uniform distribution
The distribution of the max is not hard. Let $Y$ be the max. Then $\Pr(Y\le y)=\frac{y^{10}}{(n-1)^{10}}$. FDifferentiate to get the density.
3h
comment Evaluting $\lim_{(x,y) \to(0,0)} x \cdot \ln{(x^2+2y^2)}$
You are welcome. I agree that it looked backwards, But negative numbers are weird, , that's why I go to positives as soon as I can.