Three cards are placed into a hat. One card has a white face on both sides. A second card has a white side and a red side and the third card has two red sides. A card is drawn out of the hat and falls to the floor and shows a red side. The other side was not seen. What is the probability that the card on the floor is the card with two red sides?
25% or 33%, not sure of which (I am pretty certain it is 33% tho). This is a variation of Restricted Choice. nitro
This is where I would resort to Bayes' Theorem : P(A|B) = (P(B|A)*P(A))/P(B) Let A = the event of drawing the card with 2 red sides Let B = the event that one side is red P(A|B) = (1*(1/3))/(1/2) = 2/3 or 66.67%
There are 2 cards that have a red side. What am I missing? 50% isn't right? Is it too obvious? Of course to me the interesting thing is not what the answer is, but that 75% of the answers given so far have to be wrong. Peace, rs7
Mr Subliminal is correct. Way to go! This is a problem with human probability decision making to resort to "wholes" instead of parts. There are 6 sides total (think of sides, not cards). 3 sides are white and 3 sides are red. Since the card on the floor is showing a red side, we can immediately eliminate 2 white sides (the card with two white sides). Among the two cards left that could be possible, we know that there are 3 red sides possible (one among the red/white and two belonging two the red/red card). Since two belong to the red/red card and only one belong to the red/white card, the probability of the card on the floor being the red/red card is 2/3'rds. This shows the human mind thinks in terms of raw frequencies instead of base rates and probabilities.
Ugh, I forgot to invert...But if I had been using common sense, I would have realized it must be p, not 1 - p. nitro
yeah, our brains suck... seems like i spend half my time fighting my brain... STUPID BRAIN!! STUPID, STUPID BRAIN!!
Of course it's 50%. It may seem like 66% is the probability to pick a red side out of three sides left, two reds and one white. However, you should not count one red side which is the other side of white/red card. That outcome is impossible because the side we see is red, not white and therefore that red on white/red is out. Thus, we have two possible outcomes, white and red, which makes the probability 50%.