A bag contains 9 identical balls, of which are 4 are blue and 6 are green. Three balls are taken out randomly from the bag after one another. Find the probability that all three balls are blue?(a) \(\frac {5}{8}\)(b) \(\frac {6}{19}\)(c) \(\frac {5}{21}\)(d) \(\frac {4}{7}\)The question was posed to me in examination.I would like to ask this question from Multiplication Theorem on Probability in chapter Probability of Mathematics – Class 12

A bag contains 9 identical balls, of which are 4 are blue and 6 are gr

1.	A bag contains 9 identical balls, of which are 4 are blue and 6 are green. Three balls are taken out randomly from the bag after one another. Find the probability that all three balls are blue?(a) \(\frac {5}{8}\)(b) \(\frac {6}{19}\)(c) \(\frac {5}{21}\)(d) \(\frac {4}{7}\)The question was posed to me in examination.I would like to ask this question from Multiplication Theorem on Probability in chapter Probability of Mathematics – Class 12
Answer» Right choice is (c) \(\frac {5}{21}\) For explanation I would say: Since6 balls are green, the probability of first ball to be DRAWN is green = \(\frac {6}{9}\) The probability of the first ball to be green, 8 balls are left and 5 out of them are green = \(\frac {5}{8}\) If the first two balls are green, then the probability of the THIRD ball to be green = \(\frac {4}{7}\) According to multiplication theorem, \(\frac {6}{9} . \frac {5}{8} . \frac {4}{7} = \frac {5}{21}\)

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