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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

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3 Answers 3

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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.

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.

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The combined probability of blue from Urn-1, red from Urn-2 is

$$\frac{4}{7} \times \frac{2}{7} = \frac{8}{49}.$$

The combined probability of red from Urn-1, blue from Urn-2 is

$$\frac{3}{7} \times \frac{5}{7} = \frac{15}{49}.$$

Therefore, given that one of the above two events occurred, the probability that the blue chip came from Urn-1 is

$$\frac{\frac{8}{49}}{\frac{8}{49} + \frac{15}{49}} = \frac{8}{23}. \tag1 $$

See also Bayes Theorem.

Let $A$ denote the event that the blue chip came from Urn-1.

Let $B$ denote the event that exactly one of the Urns produced a blue chip.

You are being asked for

$$p(A|B) = \frac{p(A,B)}{p(B)}.$$

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Yes, you want $$\mathrm P(U_1\mid A)=\dfrac{\mathsf P(A\mid U_1)\mathsf P(U_1)}{\mathsf P(A\mid U_1)\mathsf P(U_1)+\mathsf P(A\mid U_2)\mathsf P(U_2)}$$

However, event $A$ is the event that exactly one blue chip is drawn from the urns.

$$A=(U_1\cap U_2^\complement)\cup(U_1^\complement\cap U_2)$$

So the probability of this happening given that blue chip is drawn from urn-1, is the probability that a red chip is drawn from urn-2; and there are two of them among the seven. $$\mathsf P(A\mid U_1)=\mathsf P(U_2^\complement)=\dfrac{2}{7}$$

So $$\begin{align}\mathrm P(U_1\mid A)&=\dfrac{\mathsf P(U_2^\complement)\mathsf P(U_1)}{\mathsf P(U_2^\complement)\mathsf P(U_1)+\mathsf P(U_1^\complement)\mathsf P(U_2)}\\&=\dfrac{\tfrac 27\tfrac 47}{\tfrac 27\tfrac 47+\tfrac 57\tfrac 37}\\&~~\vdots\end{align}$$

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