2 votes
Accepted

coin flipping competition

The first solution is wrong (because it ignores ties), but is also useful. Given that there are no ties, the probability of $X<Y<Z$ is indeed $1/6$ by symmetry. And so if the probability of ...
mjqxxxx's user avatar
  • 41.5k
2 votes
Accepted

Why is a binomial distribution bimodal if p(n+1) is an integer?

If $$X \sim \operatorname{Binomial}(n,p),$$ then for $x \in \{0, \ldots, n-1\}$, $$\frac{\Pr[X = x+1]}{\Pr[X = x]} = \frac{\binom{n}{x+1} p^{x+1} (1-p)^{n-x-1}}{\binom{n}{x} p^x (1-p)^{n-x}} = \frac{n-...
heropup's user avatar
  • 137k
1 vote
Accepted

Confidence Interval for Expected Value of Binomial Distribution

It appears your GC is using the formula $$\left(\dfrac{n}{n-1}\right)\overline{p}\left(1-\overline{p}\right)$$ for its unbiased estimator of the variance. The simpler alternative, $$\overline{p}\left(...
AOS's user avatar
  • 191
1 vote
Accepted

Putting $k$ balls in n urns, which are grouped into $r$ sets.

Denote $R=2r+1$ and $m = \frac{n}{R}$ and $B={R \choose r+1}$. Let's do inclusion-exclusion with the sets $$ B_A = \{\text{all urns in groups in A are empty}\}, \text{ for } A \subset [R] \text{ with ...
ploosu2's user avatar
  • 8,886
1 vote

q 82 of Random Variable from Sheldon Ross book

There are $C(20,4)$, or $\binom{20}{4}$ different samples of four that the inspector could choose. I believe you're asking why the uncertainty about which four are chosen wasn't considered in the ...
user469053's user avatar
  • 2,137

Only top scored, non community-wiki answers of a minimum length are eligible