So I'm having problems with this one. It says: We draw with replacement two balls from an urn containing k black balls and l white balls where $k\ne l$ (with replacement the question means: draw one ball put it back in the urn, draw another ball put it back in the urn. So the experiment we are working here consists in drawing these 2 balls and comparing them, if they have the same color we start over with another experiment). This process is repeated until, for the first time, we draw two balls with different colors. Calculate the probability that the last ball you draw from this pair is white.
And the possible answers are: a) l/(k+l), b) k/(k+l), c) 1/2, d) 3/4, e) l(1-l)/(l(1-l)+k(1-k))
Should I use Bayes? I'm having problems modeling this problem.
Well I'm bad at math and I have issues to understand the questions.
So indeed as N. F. Taussig said what the problem means with the reposition part is that you look at one ball then place it back in the urn, look at another ball place it back, do this until you draw 2 balls with different colors.
So the asnwer should be obvious but i need to prove it somehow, my teacher wants some logic in the answer so I tried to solve this using Bayes.
Let P(B) and P(W) be the probability of drawing a black and a white ball respectively and, for example, inside the experiment of drawing 2 balls, P(W1) is the probability of drawing a white ball in the first draw. Let A be the event where we draw 2 balls with different colors. So we want:
$$P(W2|A) = \frac{P(A|W2)*P(W2)}{P(A)}$$
Now $P(A) = P(W1)*P(B2) + P(B1)*P(W2)$
And $P(A|W2) = P(B1W2 + W1B2|W2) = P(B1)$
Which leave us with: $$P(W2|A)=\frac{P(B1)*P(W2)}{P(B1)*P(W2)+P(W1)*P(B2)} = \frac12$$