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Gate-2018-Mechanical-engineering-->View question

Four red balls, four green balls and four blue balls are put in a box. Three balls are pulled out of the box at random one after another without replacement. The probability that all the three balls are red is (gate 2018 mechanical engineering)

Four red balls, four green balls and four blue balls are put in a box. Three balls are pulled outof the box at random one after another without replacement. The probability that all the threeballs are red is(A) 1/72 (B) 1/55 (C) 1/36 (D) 1/27

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Ans is (C)

Probability that the first ball is red =  4/12  (4 red balls available and total 12 balls) = P(A) ( say)
Probability that the second ball is red = 3/11 ( 3 red balls available and total 11 balls) = P(B) ( say)
Probability that the second ball is red = 2/10 ( 2 red balls available and total 10 balls) = P(C) ( say)
Required probability = P(A) * P(B)*P(C) = 4/12 * 3/11 * 2/10
Ans is 1/55

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