To specialists in the theory of probability. I have a portfolio of 10 stocks. What is the probability that 2 of my 10 companies will go bankrupt next year?

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Last year, 50 out of 5,000 companies went bankrupt in the US market. So the probability of a company going bankrupt is 1/100.

I have a portfolio of 10 stocks.

What is the probability that 1 of my 10 companies will go bankrupt in a year? It's easy to calculate.
The probability of one company going bankrupt is 1/100. And we take 10 companies, so we increase the odds of the event occurring by a factor of 10.
So we get the probability: 1/100 * 10 = 1/10.

What is the probability that 2 of my 10 companies go bankrupt in one year? How do we calculate this?
 
100%
 
Let's make some serious choices
 
igrok333:
let's get serious.

0.005

 
Aleksey Sergan:

0.005

What formula is used to calculate this?
 
igrok333:
Which formula is used to calculate this?

calculated as the sum of probabilities of simultaneous bankruptcies of all possible combinations of companies. the probability of simultaneous bankruptcy of two arbitrary companies is 0.01*0.01. for company 1 is 9*0.01*0.01, for company 2 is 8*0.01*0.01, etc. in total ( 9+8+7+6+5+4+3+2+1)*0.01*0.01

is true if the events are independent.

 

2\10*0.01 = 0.002 - the probability that at least one company out of 10 will go bankrupt

1\9*0.01 = 0.0011 - the probability of the second company out of 10 will go bankrupt under the condition that the first one has already gone bankrupt

0.002 * 0.0011 = 0.0000022 probability that both companies will go bankrupt

may be wrong )
  
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 doesn't work, but it's close.
 
I found a mistake. the correct answer is( 9+8+7+6+5+4+3+2+1)*0.01*0.01 = 0.0045. I checked the above where I got 0.005 in excel, rounding was correct.
 

If I understand correctly, this is a binomial distribution problem. There will be some difference between "exactly one" and "at least one"

exactly one: 0.09135172

at least one: 0.09561792

exactly two: 0.004152351

at least two: 0.0042662

Code in R:

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

dbinom(2,10,0.01)
1-pbinom(1,10,0.01)
 
igrok333:
Last year 50 out of 5,000 companies went bankrupt in the US market. So the probability of a company going bankrupt is 1/100.

I have a portfolio of 10 stocks.

What is the probability of 1 out of my 10 companies going bankrupt in a year? It's easy to calculate.
The probability of one company going bankrupt is 1/100. And we take 10 companies, so we increase the odds of the event occurring by a factor of 10.
So we get the probability: 1/100 * 10 = 1/10.

What is the probability that 2 of my 10 companies go bankrupt in one year? How do we calculate this?

The hypergeometric probability formula must be applied here.

The probability of bankruptcy is exactly 1 in 10 companies:

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

The probability of bankruptcy is exactly 2 out of 10 companies:

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

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