Monday

September 15, 2014

September 15, 2014

Posted by **Ron** on Monday, May 16, 2011 at 7:45pm.

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

- pre-calculus -
**Reiny**, Monday, May 16, 2011 at 8:13pmA) 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

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