Permutations

[thelazyman]

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.

 

[pka]

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.

 

[thelazyman]

But using the permutations formula, how can i develop a formula using that information???

 

[pka]

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.

 

[thelazyman]

OOOOOOOOOOOOOO. OKAY THANK YOU SIR.

 

[soroban]

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.

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

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

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