precalculus
posted by Ron on .
An urn contains 7 red marbles labeled (1,2,3,4,5,6,7) and 5 green marbles labeled (1,2,3,4,5) four marbles are pulled out at once. what is the probability.
A. all four marbles are red
B. more of the marbles are green than red
C. both red and green marbles are present
D. two of the marbles chosen are both labeled 5

A) prob (4reds) = (7/12)(6/11)(5/10)(4/9) = 7/99
or
Prob = C(7,4)/C(12,4) = 35/495 = 7/99
B) could be GGGG, GGGR
Prob = C(5,4)/C(12,4) + C(5,3)*C(7,1)/C(12,4
= 5/495 + 70/495 = 75/495 = 5/33
C) can't have all red or all green
prob = 1  (all red + all green)
number of all green = C(5,4) = 5
so prob = 1  (7/99 + 5/495) = 91/99
D) both are labelled 5
prob = C(2,2)*C(10,2)/C(10,4) = 45/495 = 1/11