1. ## conditional probability

a box contains 3 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?

2. let A = a green card is observed and let B= card with 2 green sides is drawn.

Find $P(B|A)$

3. That makes sense, but I am confused on what P(A intersect B) is

4. Originally Posted by holly123
a box contains 3 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?
Let $R$ be the event that the selected card is red on both sides.
Let $G$ be the event that the selected card is green on both sides.
Let $B$ be the event that the selected card is red on one and green on the other side.
Let $A$ be the event that the selected card shows green on up side.

You need to find $P(G|A)=\dfrac{P(A|G)P(G)}{ P(A|R)P(R)+ P(A|B)P(B)+ P(A|G)P(G)}$.

If you do that correctly, the answer should surprise you.

5. EDIT: Wrong attempt. See post below

6. Hello, holly123!

This is a classic trick question.
Here is an elementary explanation.

A box contains 3 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?

The "obvious" answer is $\frac{1}{2}$
. . (The other side is either Red or Green.)
However, this is wrong . . .

Let's label the six faces of the cards:

. . $\begin{array}{ccc}
\text{Card 1} & \text{Card 2} & \text{Card 3} \\
\boxed{R_1,R_2} & \boxed{G_1,G_2} & \boxed{R_3,G_3} \end{array}$

We draw one card and see a Green face.
. . (Obviously, we did not draw Card 1.)

There are three possible scenarios:

(1) You drew Card 2.
. . .You are looking at $G_1$; the other side is $G_2.$

(2) You drew Card 2.
. . .You are looking at $G_2$; the other side is $G_1.$

(3) You drew Card 3.
. . .You are looking at $G_3$; the other side is $R_3.$

In two out of the three cases, the other side is Green.

Therefore: . $P(\text{other side is Green}) \;=\;\dfrac{2}{3}$

7. Originally Posted by Soroban
Hello, holly123!

This is a classic trick question.
Here is an elementary explanation.

The "obvious" answer is $\frac{1}{2}$
. . (The other side is either Red or Green.)
However, this is wrong . . .

Let's label the six faces of the cards:

. . $\begin{array}{ccc}
\text{Card 1} & \text{Card 2} & \text{Card 3} \\
\boxed{R_1,R_2} & \boxed{G_1,G_2} & \boxed{R_3,G_3} \end{array}$

We draw one card and see a Green face.
. . (Obviously, we did not draw Card 1.)

There are three possible scenarios:

(1) You drew Card 2.
. . .You are looking at $G_1$; the other side is $G_2.$

(2) You drew Card 2.
. . .You are looking at $G_2$; the other side is $G_1.$

(3) You drew Card 3.
. . .You are looking at $G_3$; the other side is $R_3.$

In two out of the three cases, the other side is Green.

Therefore: . $P(\text{other side is Green}) \;=\;\dfrac{2}{3}$
Nice, I did't see it that way. I stad corrected.