Answer by Graham Kemp for Conditional Probability - Urn Problem
If events A and B are conditionally independent over event C (and its complement), then that and the Law of Total Probability states: $$\mathsf P(A, B)=\mathsf P(A\mid C)\mathsf P(B\mid C)\mathsf...
View ArticleAnswer by Vishu for Conditional Probability - Urn Problem
Let $A_4$ define the event that Urn $2$ is selected. Then, using Bayes’s Theorem, we have$$P(A_3 | A_1) = \frac{P(A_3)\cdot P(A_1 | A_3)}{P(A_3) \cdot P(A_1 | A_3) + P(A_4) \cdot P(A_1 | A_4)}...
View ArticleAnswer by user for Conditional Probability - Urn Problem
I do not think your answer for (ii) is correct. It is given that the urn $U_1$ is selected. Therefore $P (A_2)=\frac35$.To approach (iii) let use Bayes theorem:$$P (U_1|A_1)=\frac {P (U_1\cap A_1)}{P...
View ArticleConditional Probability - Urn Problem
Given are two urns, urn $U_1$ contains $3$ black and $2$ white balls, and urn $U_2$ contains $2$ black and $3$ white balls. A fair coin is flipped to decide which urn we should draw from. We draw $2$...
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