Even numbers from 1-10 2, 4, 6, 8, and 10 so 5 even number cards per suit= 20 even number cards that are possible. So half the deck is red because 2 of the 4 suits are so 52/2=26 possible red cards. But based on your question I am assuming the card you pick cannot be red and even at the same time. Meaning have to take out the ten cards from the 20 possible even number ones because half could be red, so that means 10 even number cards are possible. Of the 26 possible red cards 10 of them will be those red even numbers so that leaves 16 possible red cards that aren't even numbers. So 16+10=26 26/52=1/2 probability
Equal the equations then rearrange so that it is set to zero.
Use the quadratic formula to solve.
I have done this (see attached workings) but cannot get any of the solutions you've provided. I have even graphed the two functions, and the points of intersection concur with my workings (see attached graph).
