A bag containing 3 red marbles 2 green ones 8 white ones and 1 purple one She grabs 3 of them what is the probability that you one at least 1 green one?

Answer 1

First, let's note that the probability of at least 1 green one, will be theprobability of 1 green one + probability of two green ones. We only have two green marbles so we need not look at probability of greater than2 green ones. Now, there are 14 marbles total. When you grab three, you want to know what is the chance 1 or 2 of the 3 is green. One way to look at it is to say mathematically that picking two marbles at once is the same as picking one marble, then picking another, without putting the first marble back. So the we can pick the first green marble in 2/14 ways. Now if we pick another marble there is only 1 green left, so the there is a 1/13 chance we will pick it. That means there is 12 in 13 that we don't And since we are picking 3 marbles, the odd the third one is green is 1/12 if the second one is not. SO the odds of it NOT being green are 11/12 So the odds of exactly 1 green is 2/14x12/13x 11/12 and the odds of 2 exactly two greens is 2/14x1/13x. Since there are only two greens, the chance that the third is green is 0, so we don't need to worry about that. (The odds the third one is not green is 1. So we could say we are multiplying by 1.) Add these together for the desired answer. (2/14 x 12/13 x 11/12)+(2/14 x 1 /13)= 11/91+1/91=12/91 Another methods works as well. There are 3C2=6 ways to pick 2 greens from 3 and we divide by 14C2=14x13x/2=6/91 since this is the number of ways of picking 2 greens from 14 total marbles Now to this we add 3C1/91=3/91 since this is the number of ways of picking 1 green from 3 divided by 91 as above. The sum is again 12/91 (This is about .131868 or 13 percent)