Errata.... Sorry for the incessant postings.
But with the original TRUNC formula, 0.0382 and 0.0382 never (!) occur
"Never say never!"
In my zeal to avoid TMI, I forgot to address a possible retort that I anticipated.
The fact is: the original TRUNC formula
can return 0.0382 and 0.0382. But only as an accident of implementation  a defect, IMHO.
To understand that, note that TRUNC(20*0.999999999999998) returns 20 (!), as does INT(20*0.999999999999998).
Obviously, they should not, since 20*0.999999999999998 should be
and is less than 20.
The misbehavior of TRUNC and INT is
not due to limitations of 64bit binary floatingpoint arithmetic. Note that 20*0.999999999999998
20
<0 returns TRUE, which confirms that 20*0.999999999999998 results in a binary value that is indeed less than 20.
Instead, the problem is: TRUNC and INT arbitrarily round their argument to 15 significant digits before performing their operation. And since the exact decimal representation of the binary result of 20*0.999999999999998 is 19.9999999999999,
60920149533194489777088165283203125, that rounds to 20. Of course, TRUNC(20) and INT(20) are 20.
(For similar reasons, 20*0.999999999999998
<20 returns FALSE (!). Excel arbitrarily rounds the left and right operands of comparison operators to 15 significant digits, just for the purpose of the comparison.)

In theory, Excel RAND might return a value as large as 12^53 (0.99999999999999989) and as small as 2^53 (0.00000000000000011102230246251565).
(I rounded the decimal approximations to 17 significant digits.)
So yes, TRUNC(2*0.0382*
(1x*2^53)0.0382,4) does return 0.0382 for x=1,2,3,4,5. But it returns 0.0381 for x>=6. In all cases, the first argument is less than 0.0382. So, it should round down to 0.0381. This is demonstrated by the table below.
Likewise, TRUNC(2*0.0382*
x*2^530.0382,4) does return 0.0382 for x=1,2,3,4,5. But it returns 0.0381 for x>=6. In all cases, the first argument is greater (closer to zero) than 0.0382. So, it should round down toward zero to 0.0381.
Moreover, the probability that the original TRUNC formula returns 0.0382 and 0.0382 is 1 / 1,801,439,850,948,198.5 (about 0.000000000000000555111512312578) for each. That is infinitesimally less than the uniform probability of each 4dp value between 0.0382 and 0.0382, which is 1/765 (about 0.00130718954248366).
Nevertheless, I should have written: with the original TRUNC formula, 0.0382 and 0.0382
almost never (!) occur.

PS..... Yes, that means that my corrected TRUNC and alternative INT formulas might incorrectly return 0.0383 and 0.0383, at least in theory. If we are worried about the infinitesimal possibility
, the prudent implementation of those formulas is:
=
MAX(0.0382, MIN(0.0382, ROUND(TRUNC((2*0.0382+0.0001)*RAND(), 4)  0.0382, 4)
))
=
MAX(0.0382, MIN(0.0382, INT((2*382+1)*RAND()  382) / 10000
))
But I do not know if the RAND in Excel 2010 and later actually returns 15*2^53 to 12^53 or 2^53 to 5*2^53. I stumbled across the problem with INT(20*0.999999999999998) in Excel 2003, which had a very different implementation of RAND.
The ubersimpler formula =RANDBETWEEN(382,382)/10000 is looking more and more attractive.

rand v randbtwn.xlsm 



 A  B  C  D  E  F  G 

1     +residual....  exact C   

2   max RAND  1.000000000000000000000000000000  1.11E16  0.999999999999999,88897769753748434595763683319091796875   

3   min RAND  0.000000000000000111022302462516  3.45E31  0.000000000000000111022302462515,65404236316680908203125   

4        

5  2^53 times....  TRUNC(C,4)  2*0.0382*(1A*2^53)0.0382  +residual....  exact C  C<0.0382  1A*2^53 

6  1  0.0382  0.0382000000000000  1.39E17  0.0381999999999999,839683795244127395562827587127685546875  TRUE  0.999999999999999,88897769753748434595763683319091796875 

7  2  0.0382  0.0382000000000000  1.39E17  0.0381999999999999,839683795244127395562827587127685546875  TRUE  0.999999999999999,7779553950749686919152736663818359375 

8  3  0.0382  0.0382000000000000  2.78E17  0.0381999999999999,7009059171659828280098736286163330078125  TRUE  0.999999999999999,66693309261245303787291049957275390625 

9  4  0.0382  0.0382000000000000  2.78E17  0.0381999999999999,7009059171659828280098736286163330078125  TRUE  0.999999999999999,555910790149937383830547332763671875 

10  5  0.0382  0.0382000000000000  4.16E17  0.0381999999999999,56212803908783826045691967010498046875  TRUE  0.999999999999999,44488848768742172978818416595458984375 

11  6  0.0381  0.0381999999999999  +4.16E17  0.0381999999999999,4233501610096936929039657115936279296875  TRUE  0.999999999999999,3338661852249060757458209991455078125 

12        

13  2^53 times....  TRUNC(C,4)  2*0.0382*A*2^530.0382  +residual....  exact C  C>0.0382  A*2^53 

14  1  0.0382  0.0382000000000000  +6.94E18  0.0381999999999999,90907273428319967933930456638336181640625  TRUE  0.000000000000000111022302462515,65404236316680908203125 

15  2  0.0382  0.0382000000000000  +1.39E17  0.0381999999999999,839683795244127395562827587127685546875  TRUE  0.000000000000000222044604925031,3080847263336181640625 

16  3  0.0382  0.0382000000000000  +2.78E17  0.0381999999999999,7009059171659828280098736286163330078125  TRUE  0.000000000000000333066907387546,96212708950042724609375 

17  4  0.0382  0.0382000000000000  +3.47E17  0.0381999999999999,63151697812691054423339664936065673828125  TRUE  0.000000000000000444089209850062,616169452667236328125 

18  5  0.0382  0.0382000000000000  +4.16E17  0.0381999999999999,56212803908783826045691967010498046875  TRUE  0.000000000000000555111512312578,27021181583404541015625 

19  6  0.0381  0.0381999999999999  4.86E17  0.0381999999999999,49273910004876597668044269084930419921875  TRUE  0.000000000000000666133814775093,9242541790008544921875 

20        

21    0.0382000000000000  +0.00E+00  0.0381999999999999,9784616733222719631157815456390380859375   

22   P(TRUNC=0.0382)  0.000000000000000555111512312578  +2.96E31  1 / 1,801,439,850,948,198.5   

23   P(0.0382)  0.001307189542483660000000000000  +0.00E+00  1 / 765   


