A box contains 8 red, 3 white, and 9 blue balls. Draw 3 at random without replacement

Question

 Compute probability that all 3 are red ii) all 3 are white iii) 2 are red and 1 is white iv) at least 1 is white

Answer 1

1) (8/20)(7/19)(6/18)=

2)(3/20)(2/19)(1/18)=

3)(8/20)(7/19)(3/18)=

4)(17/20)(16/19)(3/18)=

 

 

1)(14/285)

2)(1/1140)

3)(7/285)

4)(816/6840)

Answer 2

A box contains 8 red, 3 white, and 9 blue balls. Draw 3 at random without replacement.

Compute probability that:

i) all 3 are red

RRR: 8/20 * 7/19 * 6/18 = 2/5 * 7/19 * 1/3 = 14/285

ii) all 3 are white

WWW: 3/20 * 2/19 * 1/18 = 1/(20*(20-1)*3)
= 1/((400-20)*3) = 1/(1200-60) = 1/1140

iii) 2 are red and 1 is white

RRW: 8/20 * 7/19 * 3/18 = 3*7*8/(20*19*18)
RWR: 8/20 * 3/19 * 7/18 = 3*7*8/(20*19*18)
WRR: 3/20 * 8/19 * 7/18 = 3*7*8/(20*19*18)
Total: 3*3*7*8/(20*19*18) = 7*8/(20*19*2) = 7/(5*19)
= 7/95

iv) at least 1 is white.

XXX: is probability of no whites
XXX: (20-3)/20 * (19-3)/19 * (18-3)/18
= 17*16*15/(20*19*18) = 34/57

P(at least 1 white) = 1 – 34/57 = 23/57

 

[Someone should check me; it's  been awhile.]