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

0.433

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) = (4)/(4+9)`

`= (4)/(13)`

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

Thus
`P(R| Q_2) = (12)/(12+5)`

`=(12)/(17)`

SO, if two coins are flipped, the probability of having two heads =
`(1)/(4)`

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

Also, the probability of having at least one single tail =
`(3)/(4)`

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

Thus: so far we knew:

`P(Q_2) = (1)/(4) \\ \\ P(Q_2) = (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) = (P(R \cap Q_2))/(P(R)) \\ \\ = (P(R|Q_2) * P(Q_2))/(P(R|Q_2) \ P(R_2) + P(R|Q_1) \ P(Q_1)) \\ \\ \\ \\ = ((12)/(17) * (1)/(4) )/((12)/(17) * (1)/(4) + (4)/(13) * (3)/(4)) \\ \\ \\ = (13)/(30)`

= 0.433