Skip to main contentSkip to solution

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}

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question