We have got 10 invitations, 4 are i blue paper, so 6 are not in blue paper.
To find the probability, we will calculate how many combinations are there and how many combination in which exactly 2 out of 5 of the invitations are blue.
The total combinations of the 5 are just 10! over 5!:
![T=(10!)/(5!)=10\cdot9\cdot8\cdot7\cdot6](https://img.qammunity.org/2023/formulas/mathematics/college/pl8g76eao183avf6nu5m2fga5h5k6m6fof.png)
The combinations of ecatly 2 blue starts with 4*3/2!, becase initially we have 4 to choose from but the second we have one less. Now we multiply by the same for the non blue, so 6*5*4/3!, and we get:
![B=(4\cdot3)/(2\cdot1)\cdot(6\cdot5\cdot4)/(3\cdot2\cdot1)=2\cdot3\cdot5\cdot4](https://img.qammunity.org/2023/formulas/mathematics/college/fn0hm6swr6174zthnl0a2i2j3ghd25repn.png)
The probability is the combinations of the event we want, B, divided by the total combinations, T:
![(B)/(T)=(2\cdot3\cdot5\cdot4)/(10\cdot9\cdot8\cdot7\cdot6)=(5\cdot4)/(10\cdot9\cdot8\cdot7)=(4)/(5\cdot9\cdot8\cdot7)=(1)/(2\cdot9\cdot2\cdot7)=(1)/(252)=0.003968\ldots\approx0.0040](https://img.qammunity.org/2023/formulas/mathematics/college/4q2reb1uyp98sidfn4jeeiale2mkpxrydz.png)
So, the probability that exactly 2 of the choosen invitations are printed in blue is 0.0040.