Probability - Showing independence of events

Question

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. Now we define two events:

$E_1:$ Urn $A$ is selected and the first marble is red.

$E_2:$ The second marble is red.

Are $E_1$ and $E_2$ independent?

I have defined three events:

$U:$ Urn $A$ is selected.

$R_1:$ The first marble is red.

$R_2:$ The second marble is red.

I know that I need to show that $\mathsf P(R_1 \cap R_2 \cap U) =$ $\mathsf P(R_1 \cap U)\mathsf P(R_2)$. I have calculated $P(R_1 \cap U) = 3/5$ and $P(R_2) = 3/5$ (since we draw from urn $A$ with replacement and there are $5$ marbles there, $3$ of which are red).

How do I continue from here? What is $\mathsf P(R_1 \cap R_2 \cap U)?$

Answer 1

I have calculated $P(R_1\cap U)=3/5$ and $P(R_2)=3/5$ (since we draw from urn $A$ with replacement and there are $5$ marbles there, $3$ of which are red).

No: $R_1\cap U$ is the event that the first marble is red and from urn A.  However, $R_2$ is just the event that the second marble is red; this does not restrict that second draw to urn A.

So $\mathsf P(R_1\cap U)$ is the probability for selecting urn A times the probability for obtaining one from the three red marbles when drawing one from five marbles in Urn A.

And $\mathsf P(R_2)$ can be found by the Law of Total Probability, or reason as you did that there are so many red marbles among the how many marbles in both urns that could be equally likely to be in a second draw.

Finally $R_1\cap U\cap R_2$ is the event that both marbles are red and drawn from urn A.  Evaluate its probability similarly.   The probability for selecting urn A times the probability for obtaining two from three red marbles when drawing two from five marbles in that urn.

Answer 2

$E_1$ and $E_2$ are not independent. The fact that your first draw was red increases the probability that your toin coss directed you to Urn $A$.

It's not correct that the probability that you selected Urn $A$ and drew a red marble is $0.6$. It's $0.3 = 0.5 \times 0.6$. The probability that both events occurred has to be less than the probability that you selected Urn $A$, and that probability is $0.5$. You provided the conditional probability of a red marble given that you selected Urn $A$, which is different.

The probability that you selected Urn $A$ is $0.5$, so the probability that you chose Urn $A$ and selected two red marbles is $0.18 = 0.5 \times 0.6 \times 0.6$.

Answer 3

So you can identify that these events are not independent simply by doing a thought experiment in the extremities, imagine if Urn A had 300 red balls and 2 blue ones, and Urn B had 2 red balls and 300 blue ones. And I picked an urn at random and picked out a red ball. Which urn would you guess I picked?

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Mathematically,

$$P(E_1|E_2) = \frac{P(E1 \cap E_2)}{P(E_2)}$$

$$=\frac{P(\text{Urn A and consecutive reds})}{P(\text{second red})}$$

$$=\frac{P(\text{Urn A and consecutive reds})}{P(\text{Second red|Urn A})P(\text{Urn A})+ P(\text{Second red|Urn A})P(\text{Urn B}) }$$ $$=\frac{(3/5)^2*(0.5)}{0.5(3/5)+0.5(2/5)}$$ $$=0.36$$ $$P(E_1)*P(E_2) = (0.5*(3/5))(0.5*2/5 + 0.5*3/5)$$ $$P(E_1)*P(E_2) = (0.5*(3/5))(0.5)$$ $$P(E_1)*P(E_2) = (0.25*(3/5))$$ $$P(E_1)*P(E_2) = 0.15$$ Hence, $$P(E_1|E_2) \neq P(E_1)P(E_2)$$