对概率论的专家。我有一个10只股票的投资组合。我的10家公司中，有2家明年会破产的概率是多少？

[删除]
100%

igrok333:

0.005

Aleksey Sergan:

0.005

igrok333:

2/10*0.01=0.002 - 10家公司中至少有一家会破产的概率

1/9*0.01=0.0011--在第一家公司已经破产的情况下，10家公司中第二家破产的概率。

0.002 * 0.0011 = 0.0000022 两家公司都会破产的概率

```void OnStart()
{
double cum = 0;
int n = 10000000;
int nk = 10;
for( int i = 0; i< n; i++ ){
for( int j = 1; j<= nk-1; j++ ){
for( int k=j+1; k<=nk; k++ ){
bool randj = MathRand()<(32767.*0.01);
bool randk = MathRand()<(32767.*0.01);
bool isfail = randj && randk;
if( isfail ) cum++;

}
}
}
double res = cum/n;
Print("res=", res );
}```

2020.01.06 13:00:57.894 fail (EURUSD,H2) res=0.0045321

0.005不起作用，但接近。

R中的代码。

```dbinom(1,10,0.01)
1-pbinom(0,10,0.01)

dbinom(2,10,0.01)
1-pbinom(1,10,0.01)```

igrok333:

P1 = (50!*4950!*10!*4990!)/(49!*9!*4941!*5000!) = (50*4950*4949*4948*4947*4946*4945*4944*4943*4942*10)/(5000*4999*4998*4997*4996*4995*4994*4993*4992*4991) = 0.09150979127569519373319974384113

P2 = (50!*4950!*10!*4990!)/(2*48!*8!*4942!*5000!) = (49*50*4950*4949*4948*4947*4946*4945*4944*4943*9*10)/(2*5000*4999*4998*4997*4996*4995*4994*4993*4992*4991) = 0.00408294394502039462124049848583