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?
Probability problem. There are 16 disks in a box.
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1"probabilistic method" has a different meaning and I removed it from the title. Also added probability tag. – 2012-04-09
6 Answers
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.
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.
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,
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 $\frac{5/16}{16/32} = 5/8$.
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.
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