A bag contains 11 balls - 5 red and 6 blue You take 3 balls at random without replacement.
What is the prob of getting 2 red and 1 blue?
solution - multiply the combinations of 2 red balls (from 5) by the combinations of 1 blue ball (from 6) and divide it by the combinations of three balls (from all 11)
answer - (5nCr2 * 6nCr1) / 11nCr3 = 10 *6 / 165 = 60/165 = 4/11
What is the prob of getting 3 balls of the same colour?
solution - there are TWO ways to get 3 balls of the same colour - 3 reds and 3 blues
so find the probablity of 3 reds add it to the probablility of 3 blues and divide this by the total combinations of any 3 balls
so add the combinations of 3 red balls (from 5) to the combinations of 3 blue balls (from 6) and divide it by the combinations of three balls (from all 11)
answer - (5nCr3 + 6nCr3) / 11nCr3 = 10 + 20 / 165 = 30/165 = 2/11
Interestingly, there is only one other possible combination - 2 blue and 1 red
The prob is - 5nCr1 * 6nCr2 / 11nCr3 = 5*15/165 = 75/165 = 5/11
so the 3 probs are 4/11 + 2/11 + 5/11 = 11/11 = 1.0 perfect !! |