Probability of all N photocopiers being broke at a time is p^N (p is probability of 1 photocopier being broken).

This brings us to quite easy formula:

p_BW^c_BW*p_C^c_C >= 1-0,999

where

p_BW is probability black-and-white is broken

c_BW is count of black-and-white photocopiers

and the same for color (C) photocopiers.

Since BW photocopiers are more reliable and there is no demand for C photocopiers we can simplify the formula to:

p_BW^c_BW >= 0,001

After putting logarithm on both sides we come up with:

c_BW * log(p_BW) <= log(0,001)

c_BW * log(0,12) >= log(0,001)

c_BW >= log(0,001) / log(0,12)

c_BW >=3,26

The solution is to buy 4 BW photocopiers (this is minimal count of photocopiers).

3 BW and 1 C photocopier would meet the criteria as well:

0,12^3*0,23=0,0004<=0,001

2 BW and 2 C photocopier would meet the criteria as well:

0,12^2*0,23^2=0,0008<=0,001

1 BW and 3 C photocopier would not meet the criteria:

0,12^1*0,23^3=0,00146004>0,001

This brings us to quite easy formula:

p_BW^c_BW*p_C^c_C >= 1-0,999

where

p_BW is probability black-and-white is broken

c_BW is count of black-and-white photocopiers

and the same for color (C) photocopiers.

Since BW photocopiers are more reliable and there is no demand for C photocopiers we can simplify the formula to:

p_BW^c_BW >= 0,001

After putting logarithm on both sides we come up with:

c_BW * log(p_BW) <= log(0,001)

c_BW * log(0,12) >= log(0,001)

c_BW >= log(0,001) / log(0,12)

c_BW >=3,26

The solution is to buy 4 BW photocopiers (this is minimal count of photocopiers).

3 BW and 1 C photocopier would meet the criteria as well:

0,12^3*0,23=0,0004<=0,001

2 BW and 2 C photocopier would meet the criteria as well:

0,12^2*0,23^2=0,0008<=0,00

1 BW and 3 C photocopier would not meet the criteria:

0,12^1*0,23^3=0,00146004>0