Probability problem -- Drawing cards with different colors on their 2 sides

[TheMathNoob]

Homework Statement

A box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability that the other side of the card is also green?

Homework Equations

P(BlA)=P(AnB)/P(A)

The Attempt at a Solution

Why intuitively the prob is not 1/2?

If you pick one card and notice the the first side is green then the prob that other side is green too would be 1/2 because there are just 2 cards that have green sides.

 

[PeroK]

Homework Statement

A box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability that the other side of the card is also green?

Homework Equations

P(BlA)=P(AnB)/P(A)

The Attempt at a Solution

Why intuitively the prob is not 1/2?

If you pick one card and notice the the first side is green then the prob that other side is green too would be 1/2 because there are just 2 cards that have green sides.

Why not do the experiment repeatedly and see what happens?

If you are able, you could write a computer program to model this: pick a card at random, pick a side at random, if it's green, then record the colour of the other side.

Or, you could model it simply by going through all the equally likely options: Card 1, side 1; Card 1, side 2 ...

 

[HallsofIvy]

There are, initially, 6 possible "sides" of these cards: R/R, R/G, G/G. Since each side of R/R or G/G is separate, to be sure of "equally likely" sides, label those R1/R2, R/G, G1/G2. If you look at one side of a card and it is green, you know it is "G of G/R" or "G1 of G1/G2" or "G2 of G2/G1". Of those three equally likely cases, the other side is also green in the "G1 of G1/G2" case or the "G2 of G2/G1" case, 2 out of the three cases.

 

[haruspex]

There are, initially, 6 possible "sides" of these cards: R/R, R/G, G/G. Since each side of R/R or G/G is separate, to be sure of "equally likely" sides, label those R1/R2, R/G, G1/G2. If you look at one side of a card and it is green, you know it is "G of G/R" or "G1 of G1/G2" or "G2 of G2/G1". Of those three equally likely cases, the other side is also green in the "G1 of G1/G2" case or the "G2 of G2/G1" case, 2 out of the three cases.

I agree, but I feel your notation is unclear.
You are right that the best way to think of it is as selecting one of six possible sides (rather than selecting a card, then selecting one side of it).
Three sides are red, three are green, so observing green leaves three equally likely possibilities for which side was selected. In only one of these is the other side red.

 

[TheMathNoob]

I agree, but I feel your notation is unclear.
You are right that the best way to think of it is as selecting one of six possible sides (rather than selecting a card, then selecting one side of it).
Three sides are red, three are green, so observing green leaves three equally likely possibilities for which side was selected. In only one of these is the other side red.

I see your intelligence and the questions that people in this forum make and I feel like the dumbest guy in this world XD. In my world, I am smart. This forum inspires my personal growth because I thought that I knew a lot, but I know nothing.

 

[Ray Vickson]

I see your intelligence and the questions that people in this forum make and I feel like the dumbest guy in this world XD. In my world, I am smart. This forum inspires my personal growth because I thought that I knew a lot, but I know nothing.

You should not be so hard on yourself. Some of your posts are very good, indeed, and some of your solutions are "spot on"---just not this time. Nobody's perfect, and we all have made (and in my case, continue to make) mistakes once in a while.