Question

Suppose that a well-mixed urn contains 100 red and blue balls, in unknown proportion. Suppose you know that there are either 80 red balls or 20 red balls.

(a) If one ball is drawn at random and seen to be red, what is your probability that the urn contains 80 red balls?

(b) If 5 balls are drawn randomly (with replacement) and each is seen to be red, what is the probability that the urn contains 80 red balls?

Answer 1

a) Given that there are 2 possibilities here.
Therefore P(80 red balls) = P(20 red balls) = 0.5

P(red ball drawn | 80 red balls) = 8/10 = 0.8
P( red ball drawn | 20 red balls) = 2/10 = 0.2

Therefore, using law of total probability, we have here:
P( red ball drawn) = P(red ball drawn | 80 red balls) P(80 red balls) +P( red ball drawn | 20 red balls) P(20 red balls)

P( red ball drawn) = 0.5*(0.8 + 0.2) = 0.5

Using Bayes theorem, probability that the urn contains 80 red balls given that a red ball is drawn is computed here as:
P( 80 red balls | red ball drawn) = 0.5*0.8 / 0.5 = 0.8

Therefore 0.8 is the required probability here.

b) As we are drawing balls with replacement here, the probability of getting a red one remains the same in each draw.

P(5 red ball drawn | 80 red balls) = 0.8⁵
P(5 red ball drawn | 20 red balls) = 0.2⁵

Using law of total probability, we have here:
P( 5 red balls drawn) = 0.5*(0.8⁵ + 0.2⁵)

Therefore using Bayes theorem now, we get the probability here as:
P( 80 red balls | 5 red balls drawn) = 0.5*0.8⁵ / 0.5*(0.8⁵ + 0.2⁵)  = 0.9990

Therefore 0.9990 is the required probability here.