Probability of picking two things

Question

There are 2 red cards and 1 blue card.

What is the probability of pulling one red card and one blue card, no specific order, only drawing 2 cards out of 3?

Logically, this would be a 66% chance right?

Since you can have (red1,blue), (red1,red2), and (red2,blue).

Now if we used combinations, this would be [(2nCr1) * (1nCr1)]/(3nCr2), which equals to 2/3.

Now I tried using fractions to multiply the probability of pulling a red card and a blue card by the following: (2/3) {the odds of pulling one red card} multiplied by (1/2) {the odds of pulling a blue card AFTER pulling a red card}.

So how come it results to 1/3?

Answer 1

Your first 2 explanations are correct. The probability is 2/3.

In your 3rd explanation using fractions, you left out the possibility of pulling the blue card first (with probability 1/3) followed by pulling a red card second after pulling a blue card (with probability 1). Multiplying these together gives 1/3, which must be added to the 1/3 you calculated of drawing red first and blue second to give the overall probability of getting one of each color as 2/3.

Another way of looking at the problem is to realize that picking 2 cards is equivalent to NOT selecting 1. Each of the 3 cards has an equal 1/3 chance of not being selected. If you don't select 1 of the 2 red cards (which happens with probability 2/3) then you must have selected one card of each color. If you don't select the blue card (which happens with probability 1/3) then you selected 2 red cards.