Answer:
0.433
Step-by-step explanation:
From the given information;
Let represent Urn 1 to be Q₁ ;
Urn 2 to be Q₂
and the event that a blue token is taken should be R
SO,
Given that:
Urn 1 comprises of 4 blue token and 9 red tokens,
Then, the probability of having a blue token | urn 1 picked is:
![P(R|Q_1) = \dfrac{4}{4+9}](https://tex.z-dn.net/?f=P%28R%7CQ_1%29%20%3D%20%5Cdfrac%7B4%7D%7B4%2B9%7D)
![= \dfrac{4}{13}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B4%7D%7B13%7D)
Urn 2 comprises of 12 blue token and 5 red tokens;
Thus ![P(R| Q_2) = \dfrac{12}{12+5}](https://tex.z-dn.net/?f=P%28R%7C%20Q_2%29%20%3D%20%5Cdfrac%7B12%7D%7B12%2B5%7D)
![=\dfrac{12}{17}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B12%7D%7B17%7D)
SO, if two coins are flipped, the probability of having two heads = ![\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B4%7D)
(since (H,H) is the only way)
Also, the probability of having at least one single tail = ![\dfrac{3}{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B4%7D)
(since (H,T), (T,H), (T,T) are the only possible outcome)
Thus: so far we knew:
![P(Q_2) = \dfrac{1}{4} \\ \\ P(Q_2) = \dfrac{3}{4}](https://tex.z-dn.net/?f=P%28Q_2%29%20%3D%20%5Cdfrac%7B1%7D%7B4%7D%20%20%5C%5C%20%5C%5C%20P%28Q_2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D)
We can now apply Naive-Bayes Theorem;
So, the probability P(of the token from Urn 2| the token is blue) = ![P(Q_2|R)](https://tex.z-dn.net/?f=P%28Q_2%7CR%29)
![P(Q_2|R) = \dfrac{P(R \cap Q_2)}{P(R)} \\ \\ = \dfrac{P(R|Q_2) * P(Q_2)}{P(R|Q_2) \ P(R_2) + P(R|Q_1) \ P(Q_1)} \\ \\ \\ \\ = \dfrac{\dfrac{12}{17} \times \dfrac{1}{4} }{\dfrac{12}{17} \times \dfrac{1}{4} + \dfrac{4}{13} \times \dfrac{3}{4}} \\ \\ \\ = \dfrac{13}{30}](https://tex.z-dn.net/?f=P%28Q_2%7CR%29%20%3D%20%5Cdfrac%7BP%28R%20%5Ccap%20Q_2%29%7D%7BP%28R%29%7D%20%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7BP%28R%7CQ_2%29%20%2A%20P%28Q_2%29%7D%7BP%28R%7CQ_2%29%20%5C%20P%28R_2%29%20%2B%20P%28R%7CQ_1%29%20%5C%20P%28Q_1%29%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7B%5Cdfrac%7B12%7D%7B17%7D%20%5Ctimes%20%5Cdfrac%7B1%7D%7B4%7D%20%7D%7B%5Cdfrac%7B12%7D%7B17%7D%20%5Ctimes%20%5Cdfrac%7B1%7D%7B4%7D%20%2B%20%5Cdfrac%7B4%7D%7B13%7D%20%5Ctimes%20%5Cdfrac%7B3%7D%7B4%7D%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%3D%20%5Cdfrac%7B13%7D%7B30%7D)
= 0.433