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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. $\frac{1}{3}$

asked in Probability by Veteran (59.9k points) | 2.4k views
0
Two balls are drawn in succession, = without replacement,

3 Answers

+18 votes
Best answer

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 $=\dfrac{1}{2}.$

answered by Loyal (8.8k points)
edited by
+1
(10C1 * 15C1)/(25C2)=0.5(hypergeometric distribution)
+5 votes
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

Answer : (c)
answered by Active (2.3k points)
+3 votes
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
answered by Active (2.4k points)
Answer:

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