A box contains three cards.
One card is red on both sides,
one card is blue on both sides, and
one card is red on one side and blue on the other.
One card is selected from the box at random, and the color on one side is observed.
If this side is red, what is the probability that the other side of the card is blue?
1/2? 1/3? Other? Please answer.
Is this how you people have fun
Bayes theorem
Are you still in school op?
1/2 if you disagree your mother is a cheap streetwhore
This is a trick question and the answer is, IT DEPENDS WHAT CARD YOU START WITH AS FIRST DRAW AND SECOND, so you can get either 1/2 or another 1/3. Enjoy.
I'm 23.
If you draw cards at random:
1/3 of the time, you'll grab the red card.
1/3 of the time, you'll grab the blue card.
1/6 of the time, you'll grab the mixed card and see its red side.
1/6 of the time, you'll grab the mixed card and see its blue side.
You're discarding any occurrence where you see a blue side. This leaves you with:
1/3 of the time, you'll grab the red card.
1/6 of the time, you'll grab the mixed card and see its red side.
You're picking the red card twice as much, so this gets normalized to 2/3 chances you get the red card, 1/3 chances you pick the mixed card. So the odds the other side is blue are 1/3.
>You're picking the red card twice as much, so this gets normalized to 2/3 chances
What
The sum of all probabilities must yield 1=100%. But since you're discarding half of them (because you saw a blue side), you need to "normalize" the shit by multiplying everything by two, so the sum is 100% again.
Hmm, what is your answer?
The graphic is deceiving:
It makes it look like you always pick the cards with that side up like in the picture, in the actual question however it is clear that when you pick a card either side can be up.
So, there are not 3 possibilities when picking a random card, but actually 6. (Each card can have 2 sides up). Of these 6 possibilities 3 make end up with you looking at a red card. Of these 3 possibilities only 1 has a blue side on the other side.
So the answer is 1/3, not 1/2 like it looks like in the picture.
Great addition to my Sup Forums bait folder. Thank you Nippon-kun.
>Sup Forums bait folder.
And people still wonder why Sup Forums is /small b/ in Spanish...
This sounds Monty Hall-ish, but I think there's something in the telling that may change it.
There are three red sides in play. The blue card can just be discarded at this point. You've just locked in one red, so now 2 of 3 remaining possibilities are also red.
2/3?
sorry, just woke up.
1/3 that it's blue.
good job pajeet.
now please use that sheet of paper to wipe your ass
Answer is 1/3
right-o
No. I have Bidet. It was just a useless page lying on by desk.
P(rr)=1/3
P(bb)=1/3
P(rb)=1/3
P(r)=0.5
P(b)=0.5
P(b|r)=p(rb)/p(r)=1/6
Where is my cookie?
Bertrand's Box Parado
/thread
(You) #
>P(b|r)=p(rb)/p(r)=(1/3)/0.5
Whoops its (1/3)/0.5=2/3 not 1/6
P(b|r)=2/3
lmao just open the box and see what card you want nerds
You missed the alternative route too.
I'm not sure if this is advanced trolling or if people legit have no idea how vhance works.
The box could have 1000 of the two blue sides card and it wouldn't matter because we are calculating a chance with a precondition.
When calculating with a precondition you divide the chance of the outcome with the chance of the precondition. In this case the chance of the precondition is 2/3 card and every card has an outcome chance of 1/3, so obviously you have a 1/2 chance to select any card that's red on the front. From there it's obvious that the other side being blue is also 1/2.