Question

An urn contains 5 red balls, 4 green balls and 4 yellow balls, for a total of 13 balls.

if five balls are randomly selected without replacement, what is the probability of selecting at least two red balls, given that at least one yellow ball is selected?

Answer 1

P(selecting at least 2 red ball and at least 1 yellow ball) = P(2 red ball, 1 yellow ball and 2 green ball) + P(3 red ball, 1 yellow ball and 1 green ball) + P(4 red ball, 1 yellow ball) + P(2 red ball, 2 yellow ball and 1 green ball) + P(2 red ball, 3 yellow ball) + P(3 red ball, 2 yellow ball)

                                                        = (5C2 * 4C1 * 4C2 / 13C5) + (5C3 * 4C1 * 4C1 / 13C5) + (5C4 * 4C1 / 13C5) + (5C2 * 4C2 * 4C1 / 13C5) + (5C2 * 4C3 / 13C5) + (5C3 * 4C2 / 13C5)

                                                        = (10*4*6/1287) + (10*4*4/1287) + (5*4/1287) + (10*6*4/1287) + (10*4/1287) + (10*6/1287)

                                                        = 760/1287

P(selecting at least one yellow ball) = 1 - P(no yellow ball) = 1 - 9C5/13C5 = 1 - 126/1287 = 1161/1287

P(selecting at least 2 red ball | selecting at least one yellow ball) = P(selecting at least 2 red ball and at least 1 yellow ball) / P(selecting at least one yellow ball)

                                                                            = (760/1287) / (1161/1287)

                                                                            = 760 / 1161

                                                                            = 0.6546