Let $(G,*)$ be a finite semigroup with identity. Prove that $G$ is a group iff $G$ has only one element $a$ such that $a^2=a$. For the right direction, since $G$ is a group with identity, then there exists $a=e_G$ i.e. $a^2 = a$. Isn't it? For the left direction, what should I do?
Let $G$ be a finite semi-group with identity such that it has only one idempotent.Is $G$ a group?  It only remains to show that for any $a\in G$ $\exists b\in G$ such that $ab=ba=e$ where $e$ is the identity of $G$  Also $e$ is the only idempotent of $G$ .How to proceed next?
How do I get rid of "U3 System" on my USB drive? I have a USB drive (SanDisk SDCZ40-016G) that mounts in a way I've never seen before. When inserted, two items appear in Places -> Computer:    "SanDisk Ultra Backup: 16GB" contains my data, and "CD Drive: U3 System" contains a Windows executable. The part with the Windows executable doesn't look useful to me so I'd like to remove it. I used GParted to delete the only partition listed on the device and then I created and formatted a new partition:    Strangely, the "CD Drive" containing the Windows executable was perfectly intact after this operation.  What's going on? I've been studying the Cox Proportional Hazards model, and  this question is glossed over in most texts.    Cox proposed fitting the coefficients of the Hazard function using a partial likelihood method, but why not just fit the coefficients of a parametric Survival function using the maximum likelihood method and a linear model?    In any cases where you have censored data, you could just find the area under the curve.  For example, if your estimate is 380 with standard deviation of 80, and a sample is censored >300, then there is an 84% probability for that sample in the likelihood calculation assuming normal error. Compute $E(X\mid X+Y)$ for independent random variables $X$ and $Y$ with standard exponential distributions Being $X$ and $Y$ independent random variables distributed as $\mathrm{Exponential}(1)$ and let $T=X+Y$ Calculate $E(X|T)$.  This is my attempt at solving this:  Being $Z=\sum_{i=1}^{n}P_{i}$ where $P_{i}$~$\mathrm{Exp}(\lambda)$ then using it's MFG you have that $$M_{P_i}(t)=\frac{\lambda}{\lambda-t}$$ then $$M_{Z}(t)=M_{\sum{P_{i}}}(t)=(M_{P_{1}})...(M_{P_{n}})=(\frac{\lambda}{\lambda-t})^n$$ with this being the MFG of a $\mathrm{Gamma}$ distribution with parameters $(n,\lambda)$  With this result is easy to see that $T$~$\mathrm{Gamma}(2,1)$  Now $E[X|T]=\int xf_{X|T}(x|t)dx$  Then using Bayes' theorem we can make $$f_{X|T}(x|t)=\frac{f_{T|X}(t|x)f_{X}(x)}{f_{T}(t)}$$ then $$f_{T|X}(t|x)=Pr(T=t|X=x)$$ $$=Pr(X+Y=t|X=x)$$ $$=Pr(x+Y=t|X=x)$$ $$=Pr(Y=t-x)=f_{Y}(t-x)$$  Using this result we can now rearrenge $$f_{X|T}(x|t)=\frac{f_{Y}(t-x)f_{X}(x)}{f_{T}(t)}$$ $$=\frac{e^{-(t-x)}.e^{-x}}{\frac{te^{-t}}{\Gamma(2)}}$$ $$=\frac{1}{t}$$ then $$E[X|T]=\int (x)(\frac{1}{t})dx$$  And this where I got stuck because $1_{x\epsilon(0,\infty})$, I made this explanation step-by-step hoping that someone can se easily check where I went wrong. 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