Question

Urn 1 contains 4 blue tokens and 9 red tokens; urn 2 contains 12 blue tokens and 5 red tokens. You flip a coin twice and if you see head two times, then you pick urn 2 else (if you see at least once the tail) you pick urn 1 and draw out a token at random from that urn. Given that the token is blue, what is the probability that the token came from urn 2

Answer 1

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:

 $P(R|Q_1) = \dfrac{4}{4+9}$

$= \dfrac{4}{13}$

Urn 2 comprises of 12 blue token and 5 red tokens;

Thus $P(R| Q_2) = \dfrac{12}{12+5}$

$=\dfrac{12}{17}$

SO, if two coins are flipped, the probability of having two heads = $\dfrac{1}{4}$

(since (H,H) is the only way)

Also, the probability of having at least one single tail = $\dfrac{3}{4}$

(since (H,T), (T,H), (T,T) are the only possible outcome)

Thus: so far we knew:

$P(Q_2) = \dfrac{1}{4} \\ \\ P(Q_2) = \dfrac{3}{4}$

We can now apply Naive-Bayes Theorem;

So, the probability P(of the token from Urn 2| the token is blue) = $P(Q_2|R)$

$P(Q_2|R) = \dfrac{P(R \cap Q_2)}{P(R)} \\ \\ = \dfrac{P(R|Q_2) * P(Q_2)}{P(R|Q_2) \ P(R_2) + P(R|Q_1) \ P(Q_1)} \\ \\ \\ \\ = \dfrac{\dfrac{12}{17} \times \dfrac{1}{4} }{\dfrac{12}{17} \times \dfrac{1}{4} + \dfrac{4}{13} \times \dfrac{3}{4}} \\ \\ \\ = \dfrac{13}{30}$

= 0.433