There are 5 blue b***s , 3 yellow b***s and 6 red b***s . suppose 3 b***s are drawn randomly each at a time without replacement. What is the probability of drawing 2 blue b***s ?
+128570 Note that we are choosing any 3 b***s from 14
And the number of total possible sets of these = C(14,3) = 364
But the only sets we're interested in are the ones where we have chosen 2 blue b***s out of 5 and then selecting 1 other ball from the remaining 9.
So the total number of these possible sets = C(5,2)*C*9,1) = 90
So the total probability = (number of sets we're interested in) / (total number of possible sets) = 90/364 = about 27.4%
+354 The first time would be a 5/14 chance. Depending on which ball you first picked will determine the chance of picking a blue ball. If a yellow ball was picked, then the chance of picking a blue ball would be 5/13. If the first time a red ball was picked, the chance would also be 5/13. If the first ball was a blue ball, then the change of picking another blue ball would be 4/13, and so on.
+128570 Note that we are choosing any 3 b***s from 14
And the number of total possible sets of these = C(14,3) = 364
But the only sets we're interested in are the ones where we have chosen 2 blue b***s out of 5 and then selecting 1 other ball from the remaining 9.
So the total number of these possible sets = C(5,2)*C*9,1) = 90
So the total probability = (number of sets we're interested in) / (total number of possible sets) = 90/364 = about 27.4%
