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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?

I think the answer should be $\frac{1}{2}$ as once the card is selected with one side green, there remain only two possibilities for the other side: either red or green.

But the answer to this question is $\frac{2}{3}$.

So,where am I wrong?Why the answer $\frac{2}{3}$ is the $\frac{1}{2}$ wrong?

Please explain! Thank you.

Silent
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6 Answers6

8

Go out from 6 sides. In 3 cases a green side shows up, and in 2 of these 3 cases the other side is green as well.

drhab
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5

You can work this out directly from the definition of conditional probability:

$$ P(G_2\mid G_1) = \frac{P(G_1 \cap G_2)}{P(G_1)} $$

Exactly one of the three cards has two green sides, so $P(G_1 \cap G_2) = 1/3$. Exactly three of the six sides that could be seen initially are green, so $P(G_1)=1/2$. Thus

\begin{align*} P(G_2\mid G_1) &= \frac{1/3}{1/2} \\ &= \frac{2}{3}. \end{align*}

dfeuer
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3

When you see a green side then it's more likely that you have the double-green card, since this has twice as many green sides as the red-green card.

Oscar Cunningham
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1

Say the cards are labelled as $C_1,C_2,C_3$ with $C_1$ being red on both sides, $C_2$ being green on both sides and $C_3$ being green on one side and red on the other. Also, let $S$ be the color of the side that was picked. Then, $$P(C_2|S=G) = \frac{P(S=G|C_2)P(C_2)}{P(S=G)} = \frac{1\times\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$

JvS
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1

The two-sided green card can be observed to have a green side in two distinct ways.

Cameron Buie
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1

Let's suppose the three cards are C1(with both sides green), C2(with one side green and another red) and C3(both sides are red). Each has two sides (mark A and B)

Now, condition is that the card selected has one side green. So, probability of C3 selection is 0.

Now, Collect the total cases with one side green:

1. C1-A
2. C1-B
3. C2-A (suppose, 'A' as green and 'B' as red)

And, favourable cases (other side should be green too) are:

1. C1-A
2. C1-B

so, probability is: 2 / 3

Vishal
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