Urn 1 has three red chips and four blue chips. Urn 2 has two red chips and five blue chips. One chip is selected at random from each urn. If exactly one of them is a blue chip, what is the probability that the chip selected from urn 1 is the blue chip?
Let A be the event that exactly one of the drawn chips is blue.
Let $U_1$ be the event that a blue chip is selected from urn 1
Let $U_2$ be the event that a blue chip is selected from urn 2.
$P(U_1|A) = \dfrac{P(A|U_1)P(U_1)}{P(A|U_1)P(U_1)+P(A|U_2)P(U_2)} = \dfrac{\dfrac{4}{7}\cdot \dfrac{4}{7}}{\dfrac{4}{7}\cdot \dfrac{4}{7} + \dfrac{5}{7}\cdot \dfrac{5}{7}}$
Is this right? It doesn't feel like it is right. I am a bit confused about defining the events and the corresponding probabilities. Please help
Best Answer
This is the issue with your attempt:
You described the event $U_1,$ “the chip selected from urn 1 is the blue chip” mentioned in the given question, as “the event that a blue chip is selected from urn 1”.
This is ambiguous because it potentially sounds like a blue chip has been selected and of relevance is whether it is from urn 1 or urn 2.
Indeed, this very misinterpretation was what subsequently tripped you up when defining $U_2,$ the complement of $U_1:$ instead of correctly defining it as the event that $\color\green{\textbf{the chip from urn 1 is not blue}},$ you defined it as the event that $\color\red{\textbf{the blue chip is from urn 2}}.$
With the corrected definition (and your already-correct application of Bayes's Theorem), you will now obtain the correct answer via $$\dfrac{\tfrac 27\tfrac 47}{\tfrac 27\tfrac 47+\tfrac 57\tfrac 37}.$$
I borrowed this fraction from Graham's answer, and it's tangential to the point I'm making: careful phrasing informs clearer thinking.