There are two Urns, Urn $A$ and Urn $B$. In Urn $A$ there are $3$ red marbles and $2$ blue ones. In Urn $B$ there are $2$ red marbles and $3$ blue ones. Through a fair coin toss we select one of the Urns and draw two marbles from it consecutively with replacement. We put each marble back after drawing it. What is the probability of the second marble being red, if the first marble we pulled is also red?
I'm not sure how to interpret this question and start my solution. Since we don't know which Urn was drawn from, does this mean that our sample space contains all $10$ marbles and the probability of the first marble being red $=$ to the probability of the second marble being red $= \frac{5}{10} = \frac{1}{2}?$ What role do the Urns play? Can someone please offer some insight? Thank you for your time.
Best Answer
Define $M_1$ and $M_2$ to be two events representing the first and the second marble being red, accordingly. We need to find the probability $P(M_2|M_1)$. Using the formula for conditional probability, we get $$P(M_2|M_1)= \frac{P(M_2\cap M_1)}{P(M_1)}.$$
Firstly consider the probability that the first marble we draw is red, i.e. denominator. It is the probability that it is red and comes from the urn A, plus the probability that it is red and comes from the urn B (this idea is called the law of total probability). More rigorously, if we define the event of drawing the marble from urn A with the set $A$ and the event of drawing the marble from urn B with the set $B$, then $A\cup B=\Omega$, $A\cap B=\emptyset$ (A and B are said to be disjoint) and $$P(M_1) = P(M_1\cap A) + P(M_1\cap B) = \frac{1}{2}\cdot\frac{3}{5} + \frac{1}{2}\cdot\frac{2}{5}= \frac{1}{2}.$$
The probability of two drawn marbles being red, i.e. numerator, is computed in the same way, $$ P(M_2\cap M_1) = P(M_2\cap M_1\cap A) + P(M_2\cap M_1\cap B) = \frac{1}{2}\cdot\left(\frac{3}{5}\right)^2 + \frac{1}{2}\cdot\left(\frac{2}{5}\right)^2 = \frac{1}{2}\cdot\frac{13}{25}.$$ Finally, $$P(M_2|M_1)=\frac{13}{25}.$$