Probability problem. There are 16 disks in a box. Question: There are 16 disks in a box. Five of them are painted red, five of them are painted blue, and six are painted red on one side, and blue on the other side. We are given a disk at random, and see that one of its sides is red. Is the other side of this disk more likely to be red or blue?
 A: There are $5 \times 2 + 6 \times 1 = 16$ red sides and of the $16$ red sides, $10$ have red on the other side, so the probability the other side is red is $\frac{10}{16}=\frac{5}{8} \gt \frac{1}{2}$, and so the other side of this disk more likely to be red.
Alternatively, if you are given a disk at random, then it is more likely to be same colour on both sides (and red and blue have equal probabilities) , so if you see a red side then the other side is more likely than not red.   
A: It may be useful to explain joriki’s comment. The other answers are all based on the assumption that the $16$ disks are equally likely to be chosen, and that once a disk has been chosen, you’re equally likely to see each side. Another way to say this is that you’re equally likely to see any of the $32$ sides. This is almost certainly the intended interpretation, but it’s not the only possible one, and the answer really does depend on the interpretation of We are given a disk at random, and see that one of its sides is red.
Suppose that the person drawing the disk at random has decided ahead of time to show you a red side if the disk has one. Then you will see a red side if and only if one of the red/red or red/blue disks was drawn. Each of these $11$ disks is equally likely to be the one that you’re shown, so on this interpretation the probability that the other side is blue is $\frac6{11}>\frac12$.
At the other extreme, the person drawing the disk at random might decide to show you a red side only if there’s no alternative. In that case the probability that the other side is blue, given that you’re shown a red side, is $0$: you must be looking at one of the red/red disks.
As you can see, the different possible interpretations affect the reduced sample space implied by your seeing a red side and consequently affect the probability.
A: Let $A$ be the random variable which can take on three values : red/red, red/blue, blue/blue (it evaluates to the pair of colors at each side of the disk that I am given). Let $B$ be the random variable that can take on two values : red or blue, and $B$ is the color of the face of the random disk that I see. Therefore, 
\begin{gather*}
P(A = \text{red/red}) = 5/16, \quad 
P(A = \text{red/blue}) = 6/16, \quad
P(A = \text{blue/blue}) = 5/16 \\
P(B = \text{red}) = 16/32 = 1/2, \quad P(B = \text{blue}) = 16/32 = 1/2.
\end{gather*}
Now the probability we wish to compute is $P(A = \text{red/red} \, | \, B = \text{red})$, and elementary probability theory shows that this probability is just 
$$
P(A = \text{red/red} \cap B = \text{red}) / P(B = \text{red}),
$$ 
and the probability on the numerator is just $P(A = \text{red/red}) = 5/16$, so that you expect a red/red disk given a red face with probability $10/16 > 1/2$. The red/red disk is therefore more probable than the red/blue disk, given a disk with a red side.
Hope that helps, 
A: The probability of $A$ given $B$ is:
$$P(A|B) = \frac{P(A \& B)}{P(B)}$$
Here, $A$ is "the other side of the disk is red" and $B$ is "the side that we can see is red".   
$P(A\&B)$ = no. of red disks / no. of disks = 5/16
$P(B)$ = no. of red sides / no. of sides = 16/32  
So the probability that the other side is red is $\frac{5/16}{16/32} = 5/8$.
A: One could use Baye's Theorem here (though, I think the "reduced sample space" approaches of the other answers are slicker).
Let $RB$ be the event that the red/blue card was chosen and let $A$ be the event that the observed side was red.
We will
 find $P(RB\mid A)$.
By Baye's Theorem:
$$\eqalign{
P(RB\mid A)
&={P(A\mid RB) P(RB)\over P(A\mid RB) P(RB)+P(A\mid RB^C) P(RB^C) }\cr
&={  (1/2)(6/16)\over   (1/2)(6/16)+ (1/2)(10/16) }\cr
&={6/16\over16/16  }\cr
&=3/8.
}
$$ 
($P(A\mid RB^C)=1/2$, since given that we did not pick the red/blue card, we picked one of the remaining 10 cards, five of which are red and the other five blue).
So the probability that we picked the red/blue card given that the observed side was red is $ 3/8<1/2$. Thus, this is the less likely outcome.
A: there are 5 red discs with red on the other side ... and 6 red discs with blue on the other side ... 
makes sense that you are more likely to find blue on the other side ... unless you lucky at blackjack
