Probability Question - using the combinations counting form C(n,k)

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Probability Question -- using the combinations counting form C(n,k)

Homework Statement



4. A bowl contains 3 red chips, 4 green chips, and 5 blue chips. A batch of 3 chips is
withdrawn at random.
(a) What is the probability that the batch contains only red chips?
(b) What is the probability that the batch contains one chip of each color?

Homework Equations





The Attempt at a Solution



Let [itex]P(N_{iR})[/itex] represent the probability of drawing a red chip on the ith turn.

Then...

[itex]P(N_{3R}) = P(N_{1R})P(N_{3R}|N_{1R}) + P(N_{2R})P(N_{3R}|N_{2R})[/itex]

Just using some simple reasoning
[itex]P(N_{1R}) = \frac{3}{12}[/itex]
[itex]P(N_{2R}|P(N_{1R})) = \frac{2}{11}[/itex]
[itex]P(N_{3R}|N_{2R}) = \frac{1}{10}[/itex]

I also know that

[itex]P(N_{2R}) = P(N_{1R})P(N_{2R}|N_{1R}) = \frac{3}{12} \frac{2}{11} = \frac{1}{22}[/itex]

So I now have

[itex]P(N_{3R}) = P(N_{1R})P(N_{3R}|N_{1R}) + P(N_{2R})P(N_{3R}|N_{2R})[/itex]
[itex]P(N_{3R}) = \frac{1}{4} P(N_{3R}|N_{1R}) + \frac{1}{22} \frac{1}{10}[/itex]
[itex]P(N_{3R}) = \frac{1}{4} P(N_{3R}|N_{1R}) + \frac{1}{220}[/itex]

I don't know how to solve for
[itex]P(N_{3R}|N_{1R})[/itex]
Do I need to use Baye's formula for this?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



4. A bowl contains 3 red chips, 4 green chips, and 5 blue chips. A batch of 3 chips is
withdrawn at random.
(a) What is the probability that the batch contains only red chips?
(b) What is the probability that the batch contains one chip of each color?

Homework Equations





The Attempt at a Solution



Let [itex]P(N_{iR})[/itex] represent the probability of drawing a red chip on the ith turn.

Then...

[itex]P(N_{3R}) = P(N_{1R})P(N_{3R}|N_{1R}) + P(N_{2R})P(N_{3R}|N_{2R})[/itex]

Just using some simple reasoning
[itex]P(N_{1R}) = \frac{3}{12}[/itex]
[itex]P(N_{2R}|P(N_{1R})) = \frac{2}{11}[/itex]
[itex]P(N_{3R}|N_{2R}) = \frac{1}{10}[/itex]

I also know that

[itex]P(N_{2R}) = P(N_{1R})P(N_{2R}|N_{1R}) = \frac{3}{12} \frac{2}{11} = \frac{1}{22}[/itex]

So I now have

[itex]P(N_{3R}) = P(N_{1R})P(N_{3R}|N_{1R}) + P(N_{2R})P(N_{3R}|N_{2R})[/itex]
[itex]P(N_{3R}) = \frac{1}{4} P(N_{3R}|N_{1R}) + \frac{1}{22} \frac{1}{10}[/itex]
[itex]P(N_{3R}) = \frac{1}{4} P(N_{3R}|N_{1R}) + \frac{1}{220}[/itex]

I don't know how to solve for
[itex]P(N_{3R}|N_{1R})[/itex]
Do I need to use Baye's formula for this?
You are doing the same thing here you did with the men and women problem. There is a simple way to solve those questions using the combinations counting form C(n,k). I don't know why you are trying to make it more complicated than that. The order in which the chips are drawn makes no difference.
 
Last edited:
  • #3
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I guess it's simply 1/C(12,3)?

I'm not sure how to solve (b) though
 
  • #4
Dick
Science Advisor
Homework Helper
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I guess it's simply 1/C(12,3)?

I'm not sure how to solve (b) though
Yes, it is. There is 1 way to pick three reds (C(3,3)) and C(12,3) total ways to choose. For b) you want to pick 1 red, 1 green and 1 blue. How many ways for each of those? Multiply them together and again divide by the total number of ways to choose.
 
  • #5
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0


Oh ok thanks I got it
 
  • #6
HallsofIvy
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I think students worry too much about formulas. There are 12 chips in the bowl, 3 of which are red. The probabability the first chip drawn is red is 3/12= 1/4. Once a red chip is drawn there are 11 chips left, 2 of which are red. The probability the second chip is red is 2/11. Once that red chip is drawn, there are 10 chips left, just 1 of which is red. The probability the third chip drawn is red is 1/10. The probability all three are red is (3/12)(2/11)(1/10). Since they are all red, there is no question of order here.

Again, the probability the first chip is red is 3/12= 1/4. There are 11 chips left, 4 of them are green. The probability the second chip is green is 4/11. There are now 10 chips left, 5 of them are blue. The probability the third chip is blue is 5/10= 1/2

The probability the chips drawn are "red, green, blue", in that order, is (3/12)(4/11)(5/10)= (1/4)(4/11)(1/2). If we did the same argument for "green, red, blue" we would get the same numerators and denominator just with the numerators in different order. That is the probability of three different colors, in any specific order, is the same. So we can get the probability of three different colors, in any order, by multiplying that result by 3!= 6, the number of ways of ordering 3 different things.
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement



4. A bowl contains 3 red chips, 4 green chips, and 5 blue chips. A batch of 3 chips is
withdrawn at random.
(a) What is the probability that the batch contains only red chips?
(b) What is the probability that the batch contains one chip of each color?

Homework Equations





The Attempt at a Solution



Let [itex]P(N_{iR})[/itex] represent the probability of drawing a red chip on the ith turn.

Then...

[itex]P(N_{3R}) = P(N_{1R})P(N_{3R}|N_{1R}) + P(N_{2R})P(N_{3R}|N_{2R})[/itex]

Just using some simple reasoning
[itex]P(N_{1R}) = \frac{3}{12}[/itex]
[itex]P(N_{2R}|P(N_{1R})) = \frac{2}{11}[/itex]
[itex]P(N_{3R}|N_{2R}) = \frac{1}{10}[/itex]

I also know that

[itex]P(N_{2R}) = P(N_{1R})P(N_{2R}|N_{1R}) = \frac{3}{12} \frac{2}{11} = \frac{1}{22}[/itex]

So I now have

[itex]P(N_{3R}) = P(N_{1R})P(N_{3R}|N_{1R}) + P(N_{2R})P(N_{3R}|N_{2R})[/itex]
[itex]P(N_{3R}) = \frac{1}{4} P(N_{3R}|N_{1R}) + \frac{1}{22} \frac{1}{10}[/itex]
[itex]P(N_{3R}) = \frac{1}{4} P(N_{3R}|N_{1R}) + \frac{1}{220}[/itex]

I don't know how to solve for
[itex]P(N_{3R}|N_{1R})[/itex]
Do I need to use Baye's formula for this?
The time to use canned formulas is later, to save time after you thoroughly understand the issues involved. Before that it is safer to proceed from first principles.

Think about the second problem this way: the event {3 different colours} = {RGB, RBG, GBR, GRB, BGR, BRG}, where those outcomes are the different colours in order they are drawn. You can easily compute P{RGB}, etc., and sum the results. However, there is a nice pattern; do you see it?
 

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