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A bag contains $10$ white balls and $15$ black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is:

1. $\frac{2}{3}$

2. $\frac{4}{5}$

3. $\frac{1}{2}$

4. $\frac{1}{3}$

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Two balls are drawn in succession, = without replacement,

Answer is $C$

probability of first ball white and second one black $=\left(\dfrac{10}{25}\right)\times \left(\dfrac{15}{24}\right)$

probability of first ball black and second one white$=\left(\dfrac{15}{25}\right)\times \left(\dfrac{10}{24}\right)$

probabilty = sum of above two probabilities $=\dfrac{1}{2}.$

edited
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(10C1 * 15C1)/(25C2)=0.5(hypergeometric distribution)
SInce 2 balls are drawn, sample space S = 2^2 = {BB,WW,BW,WB}

Probability that we get one of balck and one of white = (BW,WB)/S = 2/4 = 1/2

Probability that both balls are black=  (15C2)/(25C2) =7/20

Probability that both balls are white=  (10C2)/(25C2) =6/40

Probability that (both balls are black) or (both balls are white)= 7/20 + 6/40 =20/40=1/2

The probability that one of them is black and the other is white is: 1- 1/2 = 1/2

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