Urn 1 contains 5 red balls and 2 black balls Urn 2 contains

Urn 1 contains 5 red balls and 2 black balls. Urn 2 contains 3 red balls and 2 black ball. Urn 3 contains 4 red balls and 3 black balls. If an urn is selected at random and a ball is drawn, find the probability that it will be red.

Solution

Let P(1), P(2), P(3) = 1/3 = the probability of an urn being chosen

R = red

Thus,

P(R) = P(1) P(R|1) + P(1) P(R|1) + P(1) P(R|1)

= (1/3)*(5/7) + (1/3)*(3/5) + (1/3)*(4/7)

= 0.628571429 [ANSWER]