what is the probability that at least 3 are red?

Question

There are 7 red, 3blue, 2green. 4 are taken randomly w/out replacement.

Answer 1

i'll assume balls

12 balls : 7 red, 3 blue, 2 green

ways to take 4 out of 12 = C(12,4) = 12!/(4!8!) = 495

ways to have exactly 3 red = C(7,3) C(5,1) = 175

ways to have exactly 4 red = C(7,4) C(5,0) = 35

175 + 35 = 210

prob of at least 3 red = 210/495 = 14/33

Answer 2

Prob(at least 3 red) = prob(3 red) + prob(4 red) =

7C3 • 5 / 12C4 + 7C4 / 12C4 =
35 • 5 / 495 + 35 / 495 =
210/495 =
14/33

Answer 3

12 balls. C(12,4) = 495 possible draws

to get 4 red, C(7,4) = 35 ways
to get 3 red, C(7,3)C(5,1) = 175 ways

so 210/495 = 42/99

Answer 4

P(at least 3 red)= 1-(P(none red)+P(1 red) +P(2 red) )
= 1 - (5C4/12C4+5C3.4C1/12C4+
5C2.4C2/12C4)

=1-(5/495+40/495+60/495)
=1-105/495
=390/495

Answer 5

Combinatorial formula:

Definition: Let b(n,k) equal the binomial coefficient that some call "n-choose-k", which is equal to n!/[k!(n-k)!], where n! = "n factorial" = n(n-1)(n-2)(n-3)... (3)(2)(1).

The answer is: [b(7,3) b(5,1) + b(7,4)]/b(12,4) = 210/495 = 0.4242.