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有意思的题目

1.关于二项分布

首先是定义:在nn次独立重复的伯努利试验中,设每次试验中事件AA发生的概率为pp。用XX表示nn次重复伯努利试验中事件AA发生的次数,随机变量X的离散概率分布即为二项分布。记为:XB(n,p)X \sim B(n,p)。则有:

P(X=k)=(nk)pk(1p)nkP(X=k)=\binom{n}{k}p^k(1-p)^{n-k}

E(X)=i=0ni(ni)pi(1p)ni=i=1n(n1i1)niipi(1p)ni=npi=1n(n1i1)pi1(1p)ni=npi=0n1(n1i)pi(1p)n1i=np(p+1p)n1=np\begin{aligned}E(X)&=\sum_{i=0}^ni\cdot\binom{n}{i}p^i(1-p)^{n-i}\\&=\sum_{i=1}^n\binom{n-1}{i-1}\cdot\frac{n}{i}\cdot i\cdot p^i(1-p)^{n-i}\\&=np\sum_{i=1}^n\binom{n-1}{i-1}p^{i-1}(1-p)^{n-i}\\&=np\sum_{i=0}^{n-1}\binom{n-1}{i}p^i(1-p)^{n-1-i}\\&=np(p+1-p)^{n-1}=np\end{aligned}

D(X)=i=0n(iE(X))2(ni)pi(1p)ni=i=0n(inp)2(ni)pi(1p)ni=i=1ni2(ni)pi(1p)ni+n2p2i=0n(ni)pi(1p)ni2npi=0n(ni)ipi(1p)ni=n2p22n2p2+i=1ni2(ni)pi(1p)ni=n2p2+npi=1ni(n1i1)pi1(1p)ni=n2p2+npi=0n1(i+1)(n1i1)pi(1p)n1i=n2p2+np(1+(n1)p)=n2p2+np+n2p2np2=np(1p)\begin{aligned}D(X)&=\sum_{i=0}^n(i-E(X))^2\binom{n}{i}p^i(1-p)^{n-i}\\&=\sum_{i=0}^n(i-np)^2\binom{n}{i}p^i(1-p)^{n-i}\\&=\sum_{i=1}^ni^2\binom{n}{i}p^i(1-p)^{n-i}+n^2p^2\sum_{i=0}^n\binom{n}{i}p^i(1-p)^{n-i}-2np\sum_{i=0}^n\binom{n}{i}i\cdot p^i(1-p)^{n-i}\\&=n^2p^2-2n^2p^2+\sum_{i=1}^ni^2\binom{n}{i}p^i(1-p)^{n-i}\\&=-n^2p^2+np\sum_{i=1}^ni\cdot \binom{n-1}{i-1}p^{i-1}(1-p)^{n-i}\\&=-n^2p^2+np\sum_{i=0}^{n-1}(i+1)\cdot \binom{n-1}{i-1}p^{i}(1-p)^{n-1-i}\\&=-n^2p^2+np\left(1+(n-1)p\right)=-n^2p^2+np+n^2p^2-np^2\\&=np(1-p)\end{aligned}

2.关于超几何分布

超几何分布是统计学上一种离散概率分布。它描述了从有限N+MN+M个物件(其中包含NN个指定种类的物件,MM个其它物品)中抽出nn个物件,成功抽出该指定种类的物件的次数(不放回)。

超几何分布中的参数是N,M,nN,M,n,上述超几何分布记作XH(n,N,M)X\sim H(n,N,M) 。则有:

P(X=k)=(Nk)(Mnk)(N+Mn)P(X=k)=\frac{\binom{N}{k}\binom{M}{n-k}}{\binom{N+M}{n}}

E(X)=i=0ni(Ni)(Mni)(N+Mn)=N(N+Mn)i=1n(N1i1)(Mni)=N(N+Mn)i=0n1(N1i)(Mn1i)=N(N+Mn)(M+N1n1)=Nn!(N+Mn)!(N+M)!(N+M1)!(n1)!(N+Mn)!=nNN+M\begin{aligned}E(X)&=\sum_{i=0}^ni\cdot \frac{\binom{N}{i}\binom{M}{n-i}}{\binom{N+M}{n}}\\&=\frac{N}{\binom{N+M}{n}}\sum_{i=1}^n\binom{N-1}{i-1}\binom{M}{n-i}\\&=\frac{N}{\binom{N+M}{n}}\sum_{i=0}^{n-1}\binom{N-1}{i}\binom{M}{n-1-i}\\&=\frac{N}{\binom{N+M}{n}}\binom{M+N-1}{n-1}\\&=N\cdot \frac{n!(N+M-n)!}{(N+M)!}\cdot\frac{(N+M-1)!}{(n-1)!(N+M-n)!}\\&=\frac{nN}{N+M}\end{aligned}

D(X)=i=0n(iE(X))2(Ni)(Mni)(N+Mn)=1(N+Mn)i=0n(inNN+M)2(Ni)(Mni)=1(N+Mn)[i=0ni2(Ni)(Mni)+(nNN+M)2i=0n(Ni)(Mni)2nNN+Mi=0ni(Ni)(Mni)]=1(N+Mn)[i=0ni2(Ni)(Mni)+(nNN+M)2(N+Mn)2nN2N+M(N+M1n1)]=1(N+Mn)i=0ni2(Ni)(Mni)+(nNn+M)22(nNn+M)2=(nNn+M)2+N(N+Mn)i=1ni(N1i1)(Mni)=(nNn+M)2+N(N+Mn)i=0n1(i+1)(N1i)(Mn1i)=(nNn+M)2+N(N+Mn)[(N+M1n1)+(N1)(N+M2n2)]=(nNn+M)2+nNN+M+nN(n1)(N1)(N+M)(N+M1)\begin{aligned}D(X)&=\sum_{i=0}^n(i-E(X))^2\frac{\binom{N}{i}\binom{M}{n-i}}{\binom{N+M}{n}}\\&=\frac{1}{\binom{N+M}{n}}\sum_{i=0}^n(i-\frac{nN}{N+M})^2\binom{N}{i}\binom{M}{n-i}\\&=\frac{1}{\binom{N+M}{n}}\left[\sum_{i=0}^{n}i^2\binom{N}{i}\binom{M}{n-i}+\left(\frac{nN}{N+M}\right)^2\sum_{i=0}^{n}\binom{N}{i}\binom{M}{n-i}-\frac{2nN}{N+M}\sum_{i=0}^{n}i\binom{N}{i}\binom{M}{n-i}\right]\\&=\frac{1}{\binom{N+M}{n}}\left[\sum_{i=0}^{n}i^2\binom{N}{i}\binom{M}{n-i}+\left(\frac{nN}{N+M}\right)^2\binom{N+M}{n}-\frac{2nN^2}{N+M}\binom{N+M-1}{n-1}\right]\\&=\frac{1}{\binom{N+M}{n}}\sum_{i=0}^ni^2\binom{N}{i}\binom{M}{n-i}+\left(\frac{nN}{n+M}\right)^2-2\left(\frac{nN}{n+M}\right)^2\\&=-\left(\frac{nN}{n+M}\right)^2+\frac{N}{\binom{N+M}{n}}\sum_{i=1}^ni\binom{N-1}{i-1}\binom{M}{n-i}\\&=-\left(\frac{nN}{n+M}\right)^2+\frac{N}{\binom{N+M}{n}}\sum_{i=0}^{n-1}(i+1)\binom{N-1}{i}\binom{M}{n-1-i}\\&=-\left(\frac{nN}{n+M}\right)^2+\frac{N}{\binom{N+M}{n}}\left[\binom{N+M-1}{n-1}+(N-1)\binom{N+M-2}{n-2}\right]\\&=-\left(\frac{nN}{n+M}\right)^2+\frac{nN}{N+M}+\frac{nN(n-1)(N-1)}{(N+M)(N+M-1)}\end{aligned}

3.一道有趣的几何题

显然D=a2+b2+c2D=a^2+b^2+c^2S=πD=(a2+b2+c2)πS=\pi D=(a^2+b^2+c^2)\pi

AFD\triangle_{AFD}AOFDAO\perp FD,则有AF2=FOFDAF^2=FO\cdot FD,且SD=12CBAFS_{D}=\frac{1}{2}CB\cdot AFSA=12CBFDS_A=\frac{1}{2}CB\cdot FDSCBO=12CBFOS_{CBO}=\frac{1}{2}CB\cdot FO

那么意味着当CBCB一定时,SDAF,SAFD,SCBOFOS_D \propto AF,S_A\propto FD,S_{CBO}\propto FO,所以SD2=SASCBOS_D^2=S_A\cdot S_{CBO}

SD=12ab,SB=12bc,SC=12acS_D=\frac{1}{2}ab,S_{B}=\frac{1}{2}bc,S_C=\frac{1}{2}ac,且CB=a2+b2,CD=b2+c2,BD=a2+c2CB=\sqrt{a^2+b^2},CD=\sqrt{b^2+c^2},BD=\sqrt{a^2+c^2}

DCB=θ\angle_{DCB}=\theta,使用余弦定理则有:

cosθ=a2+b2+b2+c2a2c22a2+b2b2+c2=b2a2+b2b2+c2sinθ=a2b2+a2c2+b2c2a2+b2b2+c2\begin{aligned}\cos \theta &= \frac{a^2+b^2+b^2+c^2-a^2-c^2}{2\sqrt{a^2+b^2}\sqrt{b^2+c^2}}= \frac{b^2}{\sqrt{a^2+b^2}\sqrt{b^2+c^2}}\\\sin \theta &= \frac{\sqrt{a^2b^2+a^2c^2+b^2c^2}}{\sqrt{a^2+b^2}\sqrt{b^2+c^2}}\end{aligned}

由正弦定理可知:

SA=12a2+b2b2+c2a2b2+a2c2+b2c2a2+b2b2+c2=12a2b2+a2c2+b2c2\begin{aligned}S_A&=\frac{1}{2}\sqrt{a^2+b^2}\sqrt{b^2+c^2}\cdot \frac{\sqrt{a^2b^2+a^2c^2+b^2c^2}}{\sqrt{a^2+b^2}\sqrt{b^2+c^2}}\\&=\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2}\end{aligned}

所以SA2=SB2+SC2+SD2=14(a2b2+a2c2+b2c2)S_A^2=S_B^2+S_C^2+S_D^2=\frac{1}{4}(a^2b^2+a^2c^2+b^2c^2),故:

SB2SA2+SC2SA2+SD2SA2=1SB3SA3+SC3SA3+SD3SA3<1\begin{aligned}\frac{S_B^2}{S_A^2}+\frac{S_C^2}{S_A^2}+\frac{S_D^2}{S_A^2}=1\\\frac{S_B^3}{S_A^3}+\frac{S_C^3}{S_A^3}+\frac{S_D^3}{S_A^3}<1\end{aligned}

③得证。

不要被题目骗了,⑤本质上是在三维空间上任取一个点(当然三维坐标均大于零),这个点和原点的连线和三条轴的夹角的余弦值的平方和为一。那么这个问题就很简单了,取M(a,b,c)M(a,b,c),令OMOMx,y,zx,y,z的夹角分别为α2,β2,γ2\alpha_2,\beta_2,\gamma_2则有:

cosα2=aa2+b2+c2cosβ2=ba2+b2+c2cosγ2=ca2+b2+c2\begin{aligned}\cos \alpha_2&=\frac{a}{\sqrt{a^2+b^2+c^2}}\\\cos\beta_2&=\frac{b}{\sqrt{a^2+b^2+c^2}}\\\cos\gamma_2&=\frac{c}{\sqrt{a^2+b^2+c^2}}\end{aligned}

所以cos2α2+cos2β2+cos2γ2=1\cos^2\alpha_2+\cos^2\beta_2+\cos^2\gamma_2=1

用⑤的结论,把MM点取在OO处,发现sinα1=cosα2,sinβ1=cosβ2,sinγ1=cosγ2\sin\alpha_1=\cos\alpha_2,\sin\beta_1=\cos\beta_2,\sin\gamma_1=\cos\gamma_2,所以:sin2α1+sin2β1+sin2γ1=1\sin^2\alpha_1+\sin^2\beta_1+\sin^2\gamma_1=1

BONUS

SBSA=sinα1SCSA=sinβ1SCSA=sinγ1\begin{aligned}\frac{S_B}{S_A}&=\sin\alpha_1\\\frac{S_C}{S_A}&=\sin\beta_1\\\frac{S_C}{S_A}&=\sin\gamma_1\end{aligned}