Question

**Jon** · March 12th 2006, 10:09 AM

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**ThePerfectHacker** · March 12th 2006, 03:45 PM

Originally Posted by Jon

A bowl contains 6 red, 4 blue and 2 green marbles. A sample of 3 marbles is drawn at random w/o replacement. Find the probability that the sample contains the following:

All Blue marbles
All Green marbles
No Red marbles
Exactly 2 red Marbles
At most 2 red marbles.

All blue:
$\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}= \frac{1}{55}$
All green:
$\frac{2}{12}\times\frac{1}{11}\times \frac{0}{10}=0$
No red:
$\frac{6}{12}\times\frac{5}{11}\times\frac{5}{10}= \frac{5}{44}$
2 Red:
$\frac{_6C_2\cdot _6C_1}{_{12}C_3}=\frac{9}{22}$
At most 2: (None, 1, or 2):
$\frac{5}{44}+\frac{18}{44}+\frac{18}{44}=\frac{41} {44}$

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