Probability help.

Question

A bag contains 6 white balls and 5 black balls. Three balls are chosen at random. What is the probability that:

1. All three balls are black.
2. At most two balls are white.
3. One black ball, one white ball and a random ball.

Answer 1

Hi Krugen,

The following is how I will solve this problem to the best of my knowledge.

1. The probability that the first marble is black will be 5/11( 5 black marbles/ 11 total number of marbles). Probability of the second marble to be black will be 4/10 ( as one has already been taken). Probability of third marble being black will be 3/9 ( as two more marbles are removed now). so your equation will be 5/11*4/10*3/9 = 2/33.
2. Here you want not more than 2 to be white in the pick of three. The probability that the first one is white is 6/11. Probability the second is also white is 5/10 ( after one white being removed). Probability that the last one is black 5/9. So 6/11*5/10*5/9 = 5/33
3. Finally, one black marble is 5/11. 1 white is 6/10. Probability that 1 marble is black or white ( 4/9+5/9). so you have 5/11*6/10*(5/9+4/9)= 3/11.

Answer 2

1. 1-st Black: 5/11, 2-nd Black: 4/10, 3-rd Black: 3/9

Probability of it happening in this order: (5/11)•(4/10)•(3/9) = 0.06 or 6%

1. 1-st Black: 5/11, 2-nd White: 6/10, 3-rd Random: 9/9

Probability of it happening in this order: (5/11)•(6/10)•(9/9) = 0.27 or 27%

1. 1-st White: 6/11, 2-nd White: 5/10, 3-rd Random: 9/9

Probability of it happening in this order: (6/11)•(5/10)•(9/9) = 0.27 or 27%

1. 1-st White: 6/11, 2-nd Random: 10/10, 3-rd Random: 9/9

Probability: 6/11 or 55%