1. The problem statement, all variables and given/known data A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orange, and 6 are green. If you select 9 pieces of candy randomly, without replacement(I don't know why it is in comma's since it is necessary for the probability), give the probability that: a) Three of the hearts are white b) Three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green. 2. Relevant equations I used a multinomial probability for part (b) and the simpler binomial probability for (a). So that (a) looks like: (n C r)(p1*p2*p3)(q1*q2*q3*q4*q5*q6) where n=9 for the 9 "trials" . r=3 for the 3 successes. p = successful probability. q = failure probability. Now, this is without replacement so the probability changes from p1 to p2 to p3 and, of course q1 to q2 ... Now b, I'm not good with the notation using LaTex. But it'd be similar to: (n!)/(nw!*nt!*np!*ny!*ng!)*(w1*w2*w3)(t1*t2)(p1)(y1)(g1*g2) where the nw,...,ng represent the number of w,...,number of green. The w1,...g2 represent the respective probabilities (without replacement) of each color. 3. The attempt at a solution The solution went as follows: a) (9 C 3)(19/52)(18/51)(17/50)(33/49)(32/48)(31/47)(30/46)(29/45)(28/44) approximately = .292 b) (9!)/(19!2!1!1!2!)(19/52)(18/51)(17/50)(10/49)(9/48)(7/47)(5/46)(6/45)(5/44) approximately = .006 To me, these make sense so I'm fairly confident in the solution method. I hope that the approximation is close to the right answer as well. If it is wrong, please let me know!