You have $4$ black and $2$ red marbles in a box. You draw $3$ marbles one at a time without replacement. The first two marbles you draw can either be (1) one black and one red, or (2) no black and two red. The third marble must be black. What is the probability this occurs?
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1Your case one doesn’t seem to describe drawing two marbles. – Brian M. Scott Aug 23 '15 at 16:22
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1The meaning of "one black and no red marble" is equal to the meaning of "two black marbles" – callculus42 Aug 23 '15 at 16:35
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Please clarify statement (1). Did you mean one black and one red? – N. F. Taussig Aug 23 '15 at 16:43
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@calculus: Not in normal English, which is why we’re asking for clarification. – Brian M. Scott Aug 23 '15 at 17:03
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@BrianM.Scott Sure a clarification is needed. But if it is not required to draw exact one black marble, then my statement is true. – callculus42 Aug 23 '15 at 17:12
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Thanks for clarifying the question. Have you made an attempt to solve the problem? Where are you stuck? – N. F. Taussig Aug 23 '15 at 17:55
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The third marble is black with probability ${2\over3}$. I don't see why it "must be black". – Christian Blatter Aug 23 '15 at 18:02
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MY GUESS IS (1C4 * 1C2 *1C3 )/3C6 – user263904 Aug 23 '15 at 20:14
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You should include any work you have done on the problem in the statement of your question. I am not sure what you were attempting to do in the numerator. In the denominator, observe that any sequence of four black and two red marbles is determined by where the red marbles are placed. Hence, there are $\binom{6}{2} = 15$ distinguishable sequences of four black and two red marbles. – N. F. Taussig Aug 24 '15 at 00:47
2 Answers
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The way I read this is that you wish to find the probability that one of the following sequences of selections without replacement occurs when three marbles are selected from four black and two red marbles: $(b, r, b)$, $(r, b, b)$, $(r, r, b)$. I also interpret the question to mean that marbles of the same color are indistinguishable. Note that the first two sequences correspond to the case of one black and one red marble being selected in the first two draws followed by a black marble in the third draw, while the final sequence corresponds to the case in which the first two marbles selected are red and a black marble is drawn third.
Let's consider the sequence $(b, r, b)$. Since four of the six marbles are black, the probability of selecting a black marble first is $4/6$. This leaves us with three black and two red marbles. Hence, the probability of selecting a red marble with the second draw is $2/5$. After these two draws, we are left with three black marbles and one red marble. Thus, the probability of selecting a black marble with the third draw is $3/4$. Hence, the probability of selecting the sequence $(b, r, b)$ is $$\frac{4}{6} \cdot \frac{2}{5} \cdot \frac{3}{4} = \frac{1}{5}$$ The probabilities of the remaining two sequences occurring may be calculated similarly.
N. F. Taussig
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If in drawing the first two marbles, by one black, one red, you mean in that particular order,
Pr = $\dfrac46\cdot\dfrac25\cdot\dfrac34$
else Pr = $\dfrac46\cdot\dfrac25\cdot2!\cdot \dfrac34$
true blue anil
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