Abag of marbles contains 5 red marbles and 6 blue marbles. if three marbles are chosen, without replacement, what is the probability that all of them are red?

I would like to make a recap of this problem, as in the prior posts there are 3 different answers for questions 3 and 4.

Have 3 red and 7 blue.

1) Pick 2. P of picking at least 1 blue?

Combinatorics: P(1b&1r)+P(2b) -> (7C1 x 3C1 + 7C2)/10C2

Combinatorics shortcut: 1-P(2r) -> 1 - 3C2/10C2

Probability: P(BR)+P(RB)+P(BB) = 7/10*3/9+ 3/10*7/9+ 7/10*6/9

14/15

2) Pick 2. P of picking 0 blue?

Combinatorics: P(2r) -> 3C2/10C2

Probability: P(RR) = 3/10*2/9

1/15

3) Pick 3. P of picking at least 2 blue?

Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3

Probability: P(BBR)+P(BRB)+P(RBB)+P(BBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8+ 7/10*6/9*5/8 = 7/40+7/40+7/40+7/24

49/60 (7/15 and 42/90 are wrong; do the math)

4) Pick 3. P of picking 2 blue?

Combinatorics: P(2b&1r) -> (7C2 x 3C1)/10C3

Probability: P(BBR)+P(BRB)+P(RBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8

21/40 (8/15 and 7/40 are wrong; do the math)

Hope this may save new visitors time.

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