# Conditional Probability without Assuming Independence

I've been working through a text on probability and have been having trouble understanding how to approach this question.
The premise is that you toss a coin. If Heads, you pick a ball from Urn 1. If Tails, you pick a ball from Urn 2.
Urn 1 has 3 red balls, 3 green balls. Urn 2 has 4 red balls, 2 green balls.
You then draw two balls from the selected urn, with replacement.

Let $$R_1$$ be the event that ball 1 is red, and $$R_2$$ the event that ball 2 is red.
By the law of total probability, I can see that P($$R_1$$) = P(R|H)P(H) + P(R|T)P(T), and the same procedure can be followed for P($$R_2$$), giving $$\frac{7}{12}$$.

However, the text then asks me to find $$P(R_2|R_1)$$ WITHOUT assuming independence. (The intention is to then prove/disprove independence). I can't see how to do this at all. Any guidance is appreciated!

• Why do you capitalize "WITHOUT"? You would never assume independence to compute $P[R_2|R_1]$. It is like asking you to "drive a car WITHOUT assuming the car can autonomously drive itself." (when would you ever assume the car can autonomously drive itself?) Just use the definition of conditional probability. Commented Jan 29, 2020 at 5:39
• I'm guessing the expression "without assuming independence" comes from the text being quoted, although that would tend to make the question more confusing than it would otherwise have been, and I expect that if the "without" were emphasised at all it would have been with italics or bolding, rather than upper casing. Commented Jan 29, 2020 at 9:45

The phrase "without assuming independence" is a redundant distraction. As Michael says in his comment, just use the standard formua $$P\left(R_1|R_2\right)=\frac{P\left(R_1\cap R_2\right)}{P\left(R_2\right)}\ .$$ If this turns out to be $$\ \frac{7}{12}\$$, the same as the unconditional probability, $$\ P\left(R_1\right)\$$, then the events $$\ R_1\$$ and $$\ R_2\$$ are independent, otherwise they're not.

• I guess what confuses me about not assuming independence is - would $P(R_1 \cap R_2)$ be directly calculated by the product rule, or do I have to further condition on which urn was chosen?
– Ari
Commented Jan 29, 2020 at 16:53
• The latter. Use the formula $\ P\left(R_1\cap R_2\left|H\right.\right)P(H)+ P\left(R_1\cap R_2\left|T\right.\right)P(T)\$, just as you did for the separate probabilities of $\ R_1\$ and $\ R_2\$. Given the way the two balls are selected, there's no good reason to expect that $\ R_1\$ and $\ R_2\$ are going to be independent, but I don't see how you can be sure that they're not until you've done sufficient calculation to tell you that $\ P\left(R_1\cap R_2\right)\ne \frac{49}{144}\$. Commented Jan 29, 2020 at 21:36