Probability Urn Group Problem

Question

my group and I are having trouble figuring out how to do this.

For some reason I have an urn that contains 10 coins. 3 of the coins are blue on one side and red on the other, 3 of the coins are blue on both sides, and the 4 remaining coins are red on both sides. If I choose one coin from my precious urn and see that one side is blue, what is the conditional probability that the other side of the coin is blue as well?

So what me and my group have is a tree with a 3/10 of drawing Blue/Blue, 3/10 Blue/red, and 2/5 Red/Red. We think the only one that matters is the Blue/Blue coin thus being 3/10 * 1/2 = 3/20 (Blue the second draw). Thank you in advance from all of us!

Answer 1

You can use conditional probability. Using the fact that you're looking for $$P(a|b)={P(a\cap b)\over P(b)}$$ Where $a$ is probability that second side is blue and

$b$ is the probability that the first side is blue.

So we have: $$P(a\cap b)= \frac 3{10}$$ and $$P(b)=\frac 6{10}$$

Using the formula we get $$\frac{\frac 3{10}}{\frac 6{10}}=\frac 12$$

Answer 2

Let $A_1$ be the event that the $1st$ side is blue, and $A_2$ be the event that the $2nd$ side is blue. Then we want to find $P(A_2\mid A_1)$, and $P(A_2\mid A_1) = \dfrac{P(A_1\cap A_2)}{P(A_1)} = \dfrac{\dfrac{3}{10}}{\dfrac{6}{10}} = \dfrac{1}{2}$

Answer 3

There are two interpretations I can think of which give different results and depends on the exact situation.

The easier interpretation is probably the correct one. We have a 3 blue/blue coins, 3 blue/red coins, and 4 red/red coins in the urn. If we choose one at random without knowledge and we see that at least one of the sides is blue, we ask the question what is the chance that both sides are blue.

There are six possible scenarios out of ten for us to have picked a coin with at least one side being blue (3/10 ways of picking blue/blue, and 3/10 ways of picking blue/red). Of those six scenarios, three of them are such that both sides are blue. As $P(A|B)=\frac{P(A\cap B)}{P(B)} = \frac{|A\cap B|}{|B|}$, you have the conditional probability equals $.5$.

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The more difficult and perhaps more interesting interpretation is if suppose we laid out the coins on the table on a random side (none, some, or all of the blue/red coins may be on the blue side), and looking at the coins how they are laying, we pick one that we see the blue side of. To answer this, you can go through a similar procedure, however taking into account each of the possible scenarios in terms of how many of the blue/red coins are blue face up. The probability in this case will be higher in favor of getting a blue/blue coin.