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GreenPrint
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Probability Question -- using the combinations counting form C(n,k)
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withdrawn at random.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?
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?
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withdrawn at random.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?