Probability: independent experiments? (part 2/3)

Thank you for answering part 1 (link lichess.org/forum/off-topic-discussion/probability-independent-experiments )

The answer is 1/4 as many (if not all) of you got it right. The events are independent, because it was with replacement.
But there will be a twist in part 3.

Now lets check a 2nd scenario:

* There is a opaque bag with 2 blue marbles and 1 red marble.
* You take one marble at random and it is blue. (Quick question: What was the probability here again?)
* Now the bag has only 2 marbles, a blue one (which blue, does it matter?) and a red one.
* You take a second marble in a row, no replacement.

Given that you took a blue marble first, what is the chance of geting a red marble now?

The chance of getting a blue marble was 2/3.
Coming back to the main question, the chance of getting a red marble is 1/2 or 50%

These are questions that any 7th standard guy can answer.

So for now the two answers above state that:

* The probability of taking a red marble after taking a blue is 1/2

Ok, I can't help to giggle for the first #3 said it was a kids' problem. And altough #4 made a huge effort with baysean theory that is commendable in itself, still the sledgehammer didn't hit the nut.

Maybe this simple link will help:

www.mathsisfun.com/data/probability-events-conditional.html

The probability of geting a red marble given that we had a blue is the multiplication of the probabilities in the "path" needed to get there.

@drbeco
I retract my post #4, I was clearly wrong and I want to avoid contributing to the confusion of others, thus I removed my faulty reasoning.

So you were looking for the overall probability of first finding a blue marble, then finding the red one.
This probability should then indeed be 1/3, because in 2 of 3 cases you first find a blue marble and if that happened in 1 of 2 cases you then find the red marble, thus (2/3)*(1/2)=1/3.

Is this the answer you were looking for?

#5
Also, on second thought, the way you phrased it "Given [!] that you took a blue marble first, what is the chance of geting a red marble now?", I don't think it was clear enough that you were looking for P(B∧R), where ∧ is the logical "and" symbol, and not P(R|B).

P(B∧R)=P(B)*P(R|B)=(2/3)*(1/2)
=1/3, while

P(R|B)=[P(B|R)*P(R)]/[P(B|R)*P(R)+P(B|¬R)*P(¬R)]
=[1*(1/3)]/[1*(1/3)+(1/2)*(2/3)]
=1/2

You're successfully demonstrated that if you word a question very badly that you'll feel justified in giggling at other people.

@Thalassokrator thanks for that #8.

Please correct me if I'm wrong, but isn't P(B|¬R)=1? Just checking with you.

But yes, the answer of the question as you pointed out, P(B^R) = 1/3

Thanks