maths

Let me express my thinking here too.
In fact, in answering question, most of the people only concentrate on the event "Which card is drawn" but oversee another event "Which face of the drawn card is facing up".
So now, let's see how many possibilities are there for this scenario as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Cardrbfaces.jpg

【此圖乃本人自製圖片，未經本人同意勿擅自連結或使用】
And here's the description:
Event I: The card with both sides red is drawn and with a red face facing up.
Event II: The card with both sides red is drawn and with the other red face facing up.
Event III: The red-black card is drawn with the red face facing up.
Event IV: The red-black card is drawn with the black face facing up.
Event V: The card with both sides black is drawn and with a black face facing up.
Event VI: The card with both sides black is drawn and with the other black face facing up.
So in reality, there are a total of 6 possibilities in this question.
And now, define the following events:
A: The upper side (facing up face) is red
B: The lower side (facing down face) is black
So the question is asking for P(B|A), i.e. the probability of the lower side is black given that the upper side is red.
By the method of conditional probability, we have:
P(B∩A)/P(A) = P(B|A)
Obviously, from the diagram of events above, we can see that P(B∩A) = 1/6 and P(A) = 1/2. Therefore, the required probability for this question is:
P(B|A) = (1/6)/(1/2) = 1/3


註: 事實上, 這一題有另一個類似的問法如下:

在一次擲兩枚毫子中, 如果已知其中一個係"head" (另一個唔開估住), 咁兩個 "head" 的 probability is what?
同樣好多人係答二分一, 因為亦係只留意"第二個毫子係 head or tail?"忽略了 "兩個毫子之中, 邊個出了 head ?"
為方便解說, 我們可以假設擲了一個一元和一個二元 coin. 如此四個可能性為:
I. 一元 head, 二元 head
II. 一元 head, 二元 tail
III. 一元 tail, 二元 head
IV. 一元 tail, 二元 tail
由此可見, 能符合 "其中一個係 head" 的可能性有三個, 即 I, II 和 III. 所以兩個 head 的機會其實只有三分之一.

0.5
bcos, "the upper side of this card is red" is a happened event, probability not to be counted.

There are 2 cards can have red faces, whole red and half-red-half-black, to have bleack on the other side, P=0.5

There are 3 identical card that only differ in colour
The both sides of one of the cards is black
Therefore we can neglect the card.
Now, there is 2 cards whose upper side is red.
Only the other side of one of them is black.
Therefore the probability that its other side is black is 1/2.