OMG im not getting this

[rgrocks]

One of three cards is painted red on both sides, one is painted black on both sides and the last one is painted red on one side and black on the other. A card is randomly placed on the table. If the side up is red, what is the probability that the other side is red.

My problem right now is learning how to translate this question into math.
3 CARDS
RR P(R)=100
BB P(R)=0
BR P(R)=50

Im trying to visualize this into a venn digarm but i cant so i think thats the wrong way to go at it.
Im pretty sure this has to do something with permutation and combination but i just dont know. can someone teach me this?

If one side of the card showing is red then that eliminates the BB card
so i would believe that would make the only choices RR and RB. making the probability that the other side of the card is red 50%. but thats just what i rationalize and am not really sure because i wasnt able to show it mathematically. I assume that the real answer skews the 50% odds because of the third card BB which i removed from play because i figure that its eliminated. can someone show me mathematically how to translate this problem and solve it

 

[arthur ohlsten]

there are sic possible choices! why 6?
visualize marking the faces of the red red and the black black card R1,R2 B1, and B2
then the six possibil outcomes are
R1,R2,B1,B2,R,or B

probability of R1= 1/6
probability of R2=1/6
Probability of R=1/6

probability of R1 or R2 knowing we have r1 or R2 or R is
[1/6+1/6]/[1/6+1/6+1/6]=
(2/6) /(3/6)
2/3 answer

Arthur

 

[rgrocks]

probability of R1 or R2 knowing we have r1 or R2 or R is
[1/6+1/6]/[1/6+1/6+1/6]=
(2/6) /(3/6)
2/3 answer

I understand the six possible outcomes however i dont understand the calculation you used to carry out. I assume this is a conditional probability question. P(R1|R2) probability of R1 assuming R2 is showing? But R2=R1=R doesnt it? i mean arent they equivalent? not like a dice where 1 isnt the same as 4? i still dont get it very well. if that is true shouldnt P(R1|R2)=P(R1|R)=P(R2|R1)=P(R2|R)=P(R|R1)=P(R|R2)? can u please show me the formula you are using

Thanks
RG

 

[Denis]

One of three cards is painted red on both sides, one is painted black on both sides and the last one is painted red on one side and black on the other. A card is randomly placed on the table. If the side up is red, what is the probability that the other side is red.

I ain't no expert at probabilities, but can the problem not be worded this way:
"One of TWO cards is painted red on both sides, the other is painted red on one side and black on the other.
Both card are on a table, both red side up. If you flip one over, what is the probability that its other side is red?"

But that makes probability 1/2, not 2/3, right?
What am I missing, Galactus?

 

[pka]

Let A be the event that a red face is showing. It should be very clear that \(\displaystyle P(A)=\frac{1}{2}\).

Let B be the event that that the card with both red faces is drawn: \(\displaystyle P(B)=\frac{1}{3}\)

We are asked to find \(\displaystyle P(B|A)=\frac{P(A|B)P(B)}{P(A)}=\frac{(1)P(B)}{P(A)}= \frac{2}{3}\).

 

[rgrocks]

Let A be the event that a red face is showing. It should be very clear that \(\displaystyle P(A)=\frac{1}{2}\).

Let B be the event that that the card with both red faces is drawn: \(\displaystyle P(B)=\frac{1}{3}\)

We are asked to find \(\displaystyle P(B|A)=\frac{P(A|B)P(B)}{P(A)}=\frac{(1)P(B)}{P(A)}= \frac{2}{3}\).

.
omg i get this!!! P(A|B) is the event that if we draw the card with both sides red the face showing is red. is 100% clever... but my question is how did u get to this conclusion immediately? what phrasing of the question lead you to this equation? how should i be thinking about this if im trying to solve this myself

thanks pka and aurthor

 

[arthur ohlsten]

there are 3 possible cases ,each with a possibility of 1/6, or total probability of 3/6
3/6 of the time you see a red card

if you are in one of these 3 cases , 2 of them satisfy's you , each with a probability of 1/6
or prob. other side red = 1/6 + 1/6 when you see a red card
{1/6+1/6} / [1/6+1/6+1/6]
2/6 / 3/6
2/3 answer

Arthur

 

[rgrocks]

if you are in one of these 3 cases , 2 of them satisfy's you , each with a probability of 1/6
or prob. other side red = 1/6 + 1/6 when you see a red card
{1/6+1/6} / [1/6+1/6+1/6]
2/6 / 3/6
2/3 answer

i dont get what u mean by

if you are in one of these 3 cases , 2 of them satisfy's you , each with a probability of 1/6

please elborate. i think i follow up to if i am in one of these 3 cases but i dont follow the rest of it