Probability math problem, thanks!!!!!

aima7

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An urn contaions 1 red ball, 1 green ball and 2 blue balls. An expriment consists of drawing 2 balls in succession from the urn subject to the following rules

i) if the first ball drawn is green, then is is put into the urn before the second ball is drawn
ii) if the first ball drawn is not green, then it is not put back into the urn before the second ball is drawn.

The color of each ball is recorded when it is drawn. If one of the balls drawn is red or green, what is the probability that a blue ball was also drawn??

the anwser is 3/4 why????
 
YESSES: r,b or b,r g,b or b,g
NOES:g,r or r,g b,b or g,g
Everybody agree
I agree but disagree with the given answer.
\(\displaystyle P(r,b)=\frac{1}{4}\frac{2}{3},~P(b,r)=\frac{2}{4} \frac{1}{3},~\)

\(\displaystyle P(g,b)=\frac{1}{4} \frac{2}{4},~P(b,g)=\frac{2}{4}\frac{1}{3},~\)

That gives \(\displaystyle \frac{5}{8}\)
 
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I also thought that, but it is 3/4, due to:
"If one of the balls drawn is red or green..."
5/8 assumes 2 blues drawn are in "the run"....but they must be skipped...
No it does not assume that.

\(\displaystyle P(g,b)=\frac{1}{4}\frac{2}{4}=\dfrac{1}{8}\) because the green ball is replaced.

The the other three cases each have probability \(\displaystyle \dfrac{1}{6}\).
 
It depends on how it is read.
I read as "what is the probability that the pair is contains one green and one blue, or one red and one blue." Here the answer is \(\displaystyle \frac{5}{8}\)

It seems that some read it as "what is the probability the pair contains a blue given that the pair contains a red or a green." Here the answer is \(\displaystyle \frac{3}{4}\)
 
An urn contaions 1 red ball, 1 green ball and 2 blue balls. An expriment consists of drawing 2 balls in succession from the urn subject to the following rules

i) if the first ball drawn is green, then is is put into the urn before the second ball is drawn
ii) if the first ball drawn is not green, then it is not put back into the urn before the second ball is drawn.

The color of each ball is recorded when it is drawn. If one of the balls drawn is red or green, what is the probability that a blue ball was also drawn??

the anwser is 3/4 why????
I agree with pka that the answer depends on how the question is construed, but the more plausble reading is that the question is asking for a conditional probability. In any case, the answer of 3/4 answers that question, and the OP would like it explained.

\(\displaystyle [1]\ P(one\ blue,\ then\ one\ blue) = \dfrac{2}{4} * \dfrac{1}{3} = \dfrac{2}{12} = \dfrac{8}{48}.\)

\(\displaystyle [2]\ P(one\ blue,\ then\ one\ red) = \dfrac{2}{4} * \dfrac{1}{3} = \dfrac{2}{12} = \dfrac{8}{48}.\)

\(\displaystyle [3]\ P(one\ blue,\ then\ one\ green) = \dfrac{2}{4} * \dfrac{1}{3} = \dfrac{2}{12} = \dfrac{8}{48}.\)

\(\displaystyle [4]\ P(one\ red,\ then\ one\ blue) = \dfrac{1}{4} * \dfrac{2}{3} = \dfrac{2}{12} = \dfrac{8}{48}.\)

\(\displaystyle [5]\ P(one\ red,\ then\ one\ red) = \dfrac{1}{4} * \dfrac{0}{3} = \dfrac{0}{12} = \dfrac{0}{48}.\)

\(\displaystyle [6]\ P(one\ red,\ then\ one\ green) = \dfrac{1}{4} * \dfrac{1}{3} = \dfrac{1}{12} = \dfrac{4}{48}.\)

\(\displaystyle [7]\ P(one\ green,\ then\ one\ blue) = \dfrac{1}{4} * \dfrac{2}{4} = \dfrac{2}{16} = \dfrac{6}{48}.\)

\(\displaystyle [8]\ P(one\ green,\ then\ one\ red) = \dfrac{1}{4} * \dfrac{1}{4} = \dfrac{1}{16} = \dfrac{3}{48}.\)

\(\displaystyle [9]\ P(one\ green,\ then\ one\ green) = \dfrac{1}{4} * \dfrac{1}{4} = \dfrac{1}{16} = \dfrac{3}{48}.\)

Given that AT LEAST one ball is red or green, both balls are not blue.

\(\displaystyle P(at\ least\ one\ ball\ is\ red\ or\ green) = 1 - \dfrac{8}{48} = \dfrac{40}{48} = \dfrac{5}{6}.\)

\(\displaystyle P(exactly\ one\ ball\ is\ blue\ and\ one\ is\ red\ or\ green) = \dfrac{8 + 8 + 8 + 6}{48} = \dfrac{30}{48} = \dfrac{5}{8}.\)

p(x|y) = P(x and y) / P(y)

\(\displaystyle P(exactly\ one\ ball\ is\ blue\ given\ at\ least\ one\ ball\ is\ either\ green\ or\ red) = \dfrac{5}{8} \div \dfrac{5}{6} =\dfrac{5}{8} * \dfrac{6}{5} = \dfrac{3}{4}.\)
 
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