Factory workers

R, 92 89 bytes

cat(t(cbind("|",apply("[<-"(matrix(runif(100)<1/scan(),10),1,,0),2,cummax),"|
")),sep="")

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Ungolfed:

m <- matrix(runif(100)<1/n,10,10)   #10x10 matrix of TRUE with probability 1/n
                                    #and FALSE with probability 1-1/n
m[1,] <- 0                          #set first row to 0
m <- apply(m,2,cummax)              #take the column cumulative maxima
m <- cbind("|",m,"|\n")             #put "|" as the first and last columns
m <- t(m)                           #transpose m for the write function
cat(m,sep="")                       #print m to stdout, separated by ""

Japt -R, 22 21 20 19 18 bytes

AÆP=®|!UöêAçTÃû|C

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---

Explanation

AÆP=®|!UöêAçTÃû|C     :Implicit input of integer U
A                      :10
 Æ                     :Map the range [0,A)
  P=                   :  Assign to P (initially the empty string)
    ®                  :    Map P
     |                 :      Bitwise OR with
      !                :      The negation of
       Uö              :      A random integer in the range [0,U)
         Ã             :    End Map
          ª            :    OR, for the first element when the above will still be an empty string (falsey)
           Aç          :    Ten times repeat
             T         :      Zero
              Ã        :End Map
               û|      :Centre pad each element with "|"
                 C     :  To length 12
                       :Implicitly join with newlines and output

JavaScript (ES6), 84 bytes

n=>[...s=2e9+''].map(j=>`|${s=s.replace(/./g,i=>i&1|Math.random()*n<!+j)}|`).join`
`

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---

Recursive version, 88 bytes

n=>(g=k=>k?`|${s=s.replace(/./g,i=>i%5|Math.random()*n<(s!=k))}|
`+g(k>>3):'')(s=5e9+'')

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How?

We start with k = s = '5000000000'.

At each iteration:

• We coerce each character i of s to a number, compute i modulo 5 -- so that the leading 5 is treated like a 0 -- and randomly perform a bitwise OR with 1 with the expected probability 1/n, except on the first iteration.

• The counter k is right-shifted by 3 bits. We stop the recursion as soon as k = 0, which gives 10 iterations.

  It is important to note that 5000000000 is slightly larger than a 32-bit integer, so it is implicitly converted to 5000000000 & 0xFFFFFFFF = 705032704 just before the first bitwise shift occurs. Hence the following steps:

   step | k
  ------+-----------
     0  | 705032704
     1  | 88129088
     2  | 11016136
     3  | 1377017
     4  | 172127
     5  | 21515
     6  | 2689
     7  | 336
     8  | 42
     9  | 5
    10  | 0