Consider an urn containing two amber balls, three blue balls, and four green bal
ID: 2977709 • Letter: C
Question
Consider an urn containing two amber balls, three blue balls, and four green balls, and suppose that one ball will be randomly removed from the urn and then a second ball will be randomly removed from the eight remaining balls. (a) What is the probability that the second ball removed will be a different color than the first ball removed, given that the first ball removed will be green? (b) What is the probability that one of the two balls removed will be green given that the two balls removed will be different colors?Explanation / Answer
When first ball removed was green, urn is left with 8 balls : 2amber,3blue,3green. now,for part a, we need to remove either amber or blue in second chance. so probability = 5/8. for part b. part b can occur in 2 ways, either green ball is drawn first or second. for first probability is = 4/9 * 5/8 for second is= 5/9 *4/8 total probability is addn of both.then find all the ways in which 2 diff balls can be drawn final ans: 0.869