Question

A box has 11 marbles, 3 of which are white and 8 of which are red. A sample of 4 marbles are selected randomly from the box without replacement. What is the probability that exactly 2 are white and 2 are red

Answer 1

14/99

Select 1 marble; the chance that it is white is 4/12. Select a 2nd marble; the chance that it is white is 3/11. Select a 3rd; the chance it is white is 2/10. Select a 4th; the chance it is red is 8/9. Select a 5th; the chance it is red is 7/8. The chance of getting this specific set of 5 marbles in this order is (4/12)×(3/11)×(2/10)×(8/9)×(7/8)=(2×7)/(11×10×9).

This specific set could occur in the permutation of 5 things taken 5 at a time where 3 are identical (white), and the other 2 are also identical (red). The formula for this is 5!/(3!2!)=10.

Combining the chance of getting white, white, white, red, red with the number of ways 3 white and 2 red could have been distributed in the draw of 5 marbles gives the answer:

{(2×7)/(11×10×9)}×10=14/99

A similar process will show that the chance of getting 5 red marbles is 7/99; 4 white and 1 red is 1/99; 2 white and 3 red is 42/99; and 1 white and 4 red is 35/99.