Permutations

thelazyman

Junior Member
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Jan 14, 2006
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Hi, im having a little trouble on this question about probabilites. I am having a lot of trouble trying to set up the question and like what variable to add in and so on. The question states, A bag contains four red, three gree, and five yellow marbles. Three marbles are drawn, one at a time, without replacement. Determing the probability that the order in which they are selected is;

a) yellow, red, green
b) yellow, green, green
c) yellow, yellow red.
 
The real difficulty is that there are 27 different orders in which the balls can be drawn.
But the good news is that many have the same probability.
P(RRR)=[(4)(3)(2)]/[(12)(11)(10)].
P(RGY)= P(YRG)= P(YGR)= P(GRY)= P(GYR)= P(RYG)=[(4)(3)(5)]/[(12)(11)(10)].
P(YRY)=P(YYR)=P(RYY)=[(5)(4)(4)]/[(12)(11)(10)]

You do the other cases.
 
But using the permutations formula, how can i develop a formula using that information???
 
thelazyman said:
But using the permutations formula, how can i develop a formula using that information???
You don't!
There is no formula for this!
You simply must consider all possible draws.
 
Hello, thelazyman!

A bag contains four red, three green, and five yellow marbles.
Three marbles are drawn, one at a time, without replacement.
Determing the probability that the order in which they are selected is;

a) yellow, red, green
b) yellow, green, green
c) yellow, yellow red
Since the order of the colors is specified, it's straight-forward.

\(\displaystyle a)\;P(YRG)\:=\:\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{3}{10}\:=\:\frac{1}{22}\)

\(\displaystyle b)\;P(YGG) \:=\:\frac{5}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}\:=\:\frac{1}{44}\)

\(\displaystyle c)\;P(YYR)\:=\:\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{4}{10}\:=\:\frac{2}{33}\)
 
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