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From a box containing 4 white and 6 black balls, 3 are transferred to another empty box. From new box a ball is drawn and it is black. What is the probability that out of 3 balls transferred 2 are white and one black ?

My approach is as follows : probability = $$\frac{(4C2*6C1)}{(10C3)}$$ = 0.3

another approach i thought was $$\frac{(4*3*6)}{(10*9*8)}$$= 0.1

The answer seems to be 1/6. What am i doing wrong?

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  • $\begingroup$ Well, you haven't actually said what you are doing, so nobody can really tell for sure what you are doing wrong. If you post your reasons for the answers you gave, then probably someone will be able to tell you why it's wrong. $\endgroup$ – David Dec 10 '14 at 12:47
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You have not taken into account that you know that one of the three balls that has been transferred is black. Thus, the other two of the nine balls that are transferred to the new box must be white. Since four of the other nine balls are white, the probability that both the other balls are white is

$$\frac{\binom{4}{2}}{\binom{9}{2}} = \frac{1}{6}$$

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Let A,B,C,D be the event when 0(W) and 3(B); 1(W) and 2(B); 2(W) and 1(B); 3(W) and 0(B) are transferred to second box. Let E be the event of drawing a black ball from the second box. P(A) = 6C3/10C3 = 1/6; P(B) = 4C1*6C2/10C3 = 1/2; P(C) = 3/10; P(D) = 4C3/10C3 = 1/30; P(A) + P(B) + P(C) + P(D) = 1/6 + 1/2 + 3/10 + 1/30 = 1 Therefore Events A, B, C, D are mutual exclusive and also exhaustive. P(E/A) = 3/3= P(E/B) = 2/3 P(E/C) = 1/3 P(E/D) = 0 Required probability P(C/E)= $$\frac{P(E/C)*P(C)}{P(E/A)*P(A) + P(E/B)*P(B) + P(E/C)*P(C) + P(E/D)*P(D)}$$ = $$\frac{(1/3)*(3/10)}{1*(1/6) + (2/3) * (1/2) + (1/3 * (3/10) + 0*(1/30)}$$ = 1/6

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