A box contains ten marbles,seven of which are black and three are red.three marbles are drawn one after the other without replacement.finf the probability of choosing

One red,one black and one red marble(in that order)
At least two marbles
At least two black marbles
At most two black marbles

2 answers

P(r,b,r) = 3/10 * 7/9 * 2/8
since you are drawing 3 marbles, I'd say that P(at least 2) = 1
P(<2b) = P(0b) + P(1b) = (3/10 * 2/9 * 1/8) + 3*(7/10 * 3/9 * 2/8)
P(≤2b) = P(<2b) + P(2b) = P(<2b) + 3(7/10 * 6/9 * 3/8)
P(≥2b) = 1 - P(<2b)
I had a long answer typed out, but realized you need to multiply combinations after subtracting from the numerator or denominator. Crap. Anyways, good luck! I need to stop lurking and return to my own work. :)