Question

Urn 1 contains 2 blue tokens and 8 red tokens; urn 2 contains 12 blue tokens and 3 red tokens. You roll a die to determine which urn to choose: if you roll a 1 or 2 you choose urn 1; if you roll a 3, 4, 5, or 6 you choose urn 2. Once the urn is chosen, you 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 1?

Answer 1

We are given the probability of Urns selections here as:
P(Urn 1) = 2/6 = 1/3,
P(Urn 2) = 1 - (1/3) = 2/3

Now we are also given here that:
P( blue | Urn 1) = 2/10 = 0.2,
P( blue | Urn 2) = 12/15 = 0.8

Using law of total probability, we have here:
P( blue ) = P( blue | Urn 1)P(Urn 1) + P(blue | Urn 2) P(Urn 2) = 0.2/3 + 0.8*2/3 = 0.6

Using Bayes theorem now we have here:
P( Urn 1 | blue) = P( blue | Urn 1)P(Urn 1) / P(blue) = (0.2/3) / 0.6 = 1/9

Therefore 1/9 = 0.1111 is the required probability here.