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 October 18th, 2013, 11:02 PM #1 Newbie   Joined: Oct 2013 Posts: 10 Thanks: 0 Basic probability 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 (i.e. with no particular order). What is the probability that (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"?
 October 19th, 2013, 04:09 PM #2 Newbie   Joined: Oct 2013 Posts: 10 Thanks: 0 Re: Basic probability bump
October 19th, 2013, 05:22 PM   #3
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Re: Basic probability

Hello, xsgx!

Quote:
 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 (i.e. with no particular order).

$\text{There are: }\:{12\choose4} \:=\:\frac{12!}{4!\,8!} \:=\:495\text{ possible outcomes.}$

Quote:
 What is the probability that: (a) all four marbles are red?

$P(\text{4R}) \:=\:\frac{{7\choose4}}{495} \:=\:\frac{35}{495} \:=\:\frac{7}{99}$

Quote:
 (b) more of the marbles are green than red?

$P(4G,\,0R) \:=\:\frac{{5\choose4}{7\choose0}}{495} \:=\:\frac{5\,\cdot\,1}{495} \:=\:\frac{5}{495}$

$P(3G,\,1R) \:=\:\frac{{5\choose3}{7\choose1}}{495} \:=\:\frac{10\,\cdot\,7}{495} \:=\:\frac{70}{495}$

$\text{Therefore: }\:P(G \,=>\.R) \:=\:\frac{5}{495}\,+\,\frac{70}{495} \:=\:\frac{75}{495} \:=\:\frac{5}{33}$

Quote:
 (c) both red and green marbles are present?

$P(\text{all R}) \:=\:\frac{35}{495}$

$P(\text{all G}) \:=\:\frac{{5\choose4}}{495} \:=\:\frac{5}{495}$

$P(\text{one color}) \:=\:\frac{5}{495}\,+\,\frac{35}{495} \:=\:\frac{40}{495} \:=\:\frac{8}{99}$

$\text{Therefore: }\:P(\text{both colors}) \;=\;1\,-\,\frac{8}{99} \;=\;\frac{91}{99}$

Quote:
 (d) two of the marbles chosen are both labeled "5"?

$\text{We already have R5 and G5 (one way).$
$\text{The other two can be any of the other 10 marbles: }\:{10\choose2} \,=\,45\text{ ways.}$

$P(\text{R5, G5}) \:=\:\frac{45}{495} \:=\:\frac{1}{11}$

 October 19th, 2013, 06:03 PM #4 Newbie   Joined: Oct 2013 Posts: 10 Thanks: 0 Re: Basic probability Wow, thank you! Just what I needed, a step by step explanation.
 October 21st, 2013, 05:18 AM #5 Newbie   Joined: Oct 2013 Posts: 9 Thanks: 0 Re: Basic probability Very neat solution. My mind did not think as laterally as you did. By thinking "aloud", I could do (a), the first part of (b) and (c) with some luck. For (a), I did 7/12 x 6/11 x 5/10 x 4/9 = =7/99 For straightforward 4 greens, I got 1/99; I was then faced with doing all the combinations of greens and red. Well done.
 October 21st, 2013, 09:31 PM #6 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Basic probability [color=#0000BF]nubian123[/color], Using your method, the probabilities of the various combinations of greens and reds are not too hard to calculate: $P_{1G,3R}=\frac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} \cdot \frac{5}{9} \cdot \frac{4!}{3! \cdot 1!}=\frac{35}{99}$ $P_{2G,2R}=\frac{5}{12} \cdot \frac{4}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{4!}{2! \cdot 2!}=\frac{42}{99}$ $P_{3G,1R}=\frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} \cdot \frac{7}{9} \cdot \frac{4!}{3! \cdot 1!}=\frac{14}{99}$ The factorial expressions multiplied at the end are the number of permutations in the order that the marbles may be picked. For example, with 1 green and 3 red; GRRR, RGRR, RRGR, RRRG gives 4. For 2 green, 2 red; RRGG, RGRG, RGGR, GRRG, GRGR, GGRR gives 6. Notice that all probabilities add to 1 as they should: $\frac{7}{99}+\frac{35}{99}+\frac{42}{99}+\frac{14} {99}+\frac{1}{99}=\frac{99}{99}$ d) is worked as: $\frac{2}{12}\cdot \frac{1}{11} \cdot \frac{10}{10} \cdot \frac{9}{9} \cdot \frac{4!}{2! \cdot 2!}=\frac{1}{11}$

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# if i have 7 red marbles and 5 green marbles what is the probability that all four marbles are red?

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