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There are two urns containing colored balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have probability 50% of being chosen) and then a ball is drawn at random from one of the two urns. If a red ball is drawn, what is the probability that it comes from the first urn?

my question is:

so here (1/2) is the red intersection first urn or it`s the p(red|first urn) I need explanation for that point please and if the second one is the right why it p(red|first urn) and not p(first urn| red)

thanks in advance

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If we define the following events:

$U1$: Urn $1$ is chosen

$U2$: Urn $2$ is chosen

$R$: a red ball is chosen

then we have:

$P(U1)=P(U2)=P(R|U1)=50$%

The $50$% is not $P(U1 \cap R)$, for we have $P(U1 \cap R) = P(R|U1) \cdot P(U1) = \frac{1}{2}\cdot \frac{1}{2}=0.25$

That is, $P(U1 \cap R)$ is the probability that you choose urn $1$ and pick a red ball, so that mreans you first need to pick urn $1$ ($50$% chance of that), and then you also need to pick a red ball, given that you picked urn $1$, and that is also $50$%, so for both to happen it's $50$% times $50$% is $25$%

Finally, you are asked to find $P(U1|R)$, and for that use Bayes' Law

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  • $\begingroup$ but you figured that out from the syntax of the problem about the 50% being intersection or conditional probability $\endgroup$ – Ahmed Araby May 16 '18 at 17:11
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    $\begingroup$ @AhmedAraby The problem says that urn 1 contains $50$ red balls and $50$ blue balls, and so from that we can say that $P(R|U1) = \frac{1}{2}$ $\endgroup$ – Bram28 May 16 '18 at 17:20
  • $\begingroup$ okay , but when we can say that is probability of R intersection U1 what is the change that will happen to the problem syntax here in order to say that is an intersection $\endgroup$ – Ahmed Araby May 16 '18 at 18:17
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    $\begingroup$ @AhmedAraby $P(R \cap U1)$ is the probability of picking urn 1 and getting a red ball. Or, in short: the probability of picking a red ball from urn $1$, given that you have not yet decided from which urn you're going to pick a ball. $\endgroup$ – Bram28 May 16 '18 at 18:28
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You have two urns with known amounts of red and blue balls.   This tells you the (conditional) probabilities for drawing a red ball from each given urn.

$$\mathsf P(R\mid U_1) = 50/100 = 5/10\\ \mathsf P(R\mid U_2)=30/100=3/10$$

You are to randomly select an urn (ie: without bias).   This tells you the (marginal) probabilities for selecting each urn. $$\mathsf P(U_1)=1/2\\ \mathsf P(U_2)=1/2$$

So then the probabilities for selecting each urn and drawing a red ball, will be the products.$$\mathsf P(U_1\cap R)= \mathsf P(U_1)~\mathsf P(R\mid U_1)= 5/20\\\mathsf P(U_2\cap R)=\mathsf P(U_2)~\mathsf P(R\mid U_2)=3/20$$

The Law of Total Probability tells us the (marginal) probability for drawing a red ball.$$\mathsf P(R)~{=\mathsf P(U_1\cap R)+\mathsf P(U_2\cap R)\\=\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)\\=8/20}$$

Now then, you are to evaluate the probability that urn 1 was selected given that a red ball was drawn.   From this we use wither the defiition for conditional probability, or Bayes' Rule.$$\begin{split}\mathsf P(U_1\mid R) &= \dfrac{\mathsf P(U_1\cap R)}{\mathsf P(R)} &\quad&=\dfrac{\mathsf P(U_1)~\mathsf P(R\mid U_1)}{\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)}\\ &=\dfrac{5}{8} \end{split}$$


Note: This should be anticipated, because 50 from the 80 red balls are in urn 1 and both urns contain an equal total count for balls.

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