0
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Commented Jan 29, 2020 at 5:39
  • $\begingroup$ 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. $\endgroup$ Commented Jan 29, 2020 at 9:45

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Ari
    Commented Jan 29, 2020 at 16:53
  • $\begingroup$ 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}\ $. $\endgroup$ Commented Jan 29, 2020 at 21:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .