Think of the SS numbers as being a collection of digits. If any of the digits (say the first one for example) are 8 then all the remaining digits can be anything but 8, meaning there are 9 choices for each remaining digit. This effectively becomes an 8 digit number with the constraint that there can only be 9 choices for a digit. This means for the first digit being 8 we have 9⁸ remaining possible numbers. Since we will do this for each of the remaining digits (they each get to be 8 then we count their 9⁸ configurations) we end up with 9(9⁸) or 9⁹≈3.8742E8
The probability is 4⋅3⋅2⋅1/10•9•8•7=(4/4)(6/0)/(10/4)=1/210 so the probability of selecting 3 red balls would be 1/210
I’m pretty sure it’s the 4th one
F(x)=y=2x+3
P (1,5)
Substitute x and y:
5=2*(1)+3
5=5 true
Point P (1,5) is on the graph of f.
