# conditional probability problem: picking chips

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

## 3 Answers

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.

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

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

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}