Two marbles are selectod randomly five a box containing 5 red, 7 white and 8 thore marbles one after the other withoet replacement. Find the probutility that (a) both are red; (b) one is white and the other is blase, (c) both are of the mane coloer

We have a box with 5 red, 7 white, and 8 blue marbles (5 + 7 + 8 = 20 total). Two marbles are drawn one after the other without replacement. (a) Probability both are red:  • The probability the first marble is red is 5/20.  • Once a red is removed, 4 red remain out of 19 total.  • So, P(both red) = (5/20) × (4/19) = 20/380 = 1/19. (b) Probability one is white and the other is blue: This can happen in two orders.  Order 1: White then blue:   • First marble white: 7/20.   • Then blue: 8/19.   • So, probability = (7/20) × (8/19).  Order 2: Blue then white:   • First marble blue: 8/20.   • Then white: 7/19.   • So, probability = (8/20) × (7/19).  Adding these:   P(white and blue in any order) = (7×8/20×19) + (8×7/20×19) = (56 + 56) / (380) = 112/380.   Simplify by dividing numerator and denominator by 4: 112/380 = 28/95. (c) Probability both marbles are of the same color: We sum the probabilities for red–red, white–white, and blue–blue.  • Red–red: As in (a): (5/20) × (4/19) = 10/190.  • White–white:   – First white: 7/20.   – Then white: 6/19.   – So, = (7/20) × (6/19) = 42/380 = 21/190.  • Blue–blue:   – First blue: 8/20.   – Then blue: 7/19.   – So, = (8/20) × (7/19) = 56/380 = 28/190.  Total probability = (10 + 21 + 28) / 190 = 59/190. Thus, the answers are:  (a) 1/19,  (b) 28/95, and  (c) 59/190.