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A bag contains 5 black, 3 white and 2 red balls. Three balls are drawn in succession. What is the probability that the first ball is red, the second ball is black and the third ball is white?

Solution

✅ Correct Option: 1

Total balls = 10. Drawing without replacement.

P(red first) = 210\frac{2}{10}

P(black second | red drawn) = 59\frac{5}{9}

P(white third | red and black drawn) = 38\frac{3}{8}

Required probability = 210×59×38=30720=124\frac{2}{10} \times \frac{5}{9} \times \frac{3}{8} = \frac{30}{720} = \frac{1}{24}

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