Probability-Combination math help requested.

Question

1. A jar contains only red, yellow and orange marbles. If there are 3 red, 5 yellow and 4 orange marbles, and 3 marbles are chosen from the jar at random without replacing any of them, what is the probability that 2 yellow, 1 red and no orange marbles will be chosen?
2. All the 12 marbles in the jar are shuffled. Seven marbles are selected. Find the probability that of the seven marbles at least four will be red and the remaining must be of both yellow and orange color.

Answer 1

1. We can pick the marbles in order, one at a time, to make it easier. Note, however, that the order in which you pick these marbles doesn't matter, as the problem does not care about the order in which you pick each color marble. It does matter that we are NOT replacing each marble we take out. This means that every time we take a marble out, the total number of marbles left in the jar goes down by 1.

The odds of picking 1 yellow marble at first is 5/12, as there are 5 yellow marbles out of the total 12 marbles in the jar. Take a yellow marble, and there are now 11 marbles left in the jar.

Now, the odds of picking another yellow marble is 4/11, as there are now only 4 yellow marbles out of the total 11 marbles left in the jar. Take a yellow marble, and there are now 10 marbles left in the jar.

With your last pick, the odds of picking a red marble is 3/10, as there are 3 red marbles out of the total 10 marbles left in the jar.

We can now multiply the odds of these 3 events together: 5/12 * 4/11 * 3/10 = 60/1320 = 1/22

2. This part is tricky. Assuming that we are still picking without replacement, then it is IMPOSSIBLE to pick 4 red marbles, as there are a total of only 3 red marbles in the jar. The probability would then be 0, and your job is done.

However, if they are asking about picking with replacement, the problem becomes more difficult.

To pick 7 marbles with replacement, with at least 4 red and the remaining having both yellow and orange, we have to calculate the probabilities of picking 4 and 5 red marbles. The reason we don't calculate for picking 6 and 7 is because those don't allow for us to pick a yellow and orange, since there wouldn't be 2 marble picks left.

With replacement

the probability of picking 4 red marbles: (3/12)⁴, or (1/4)⁴, which is 1/256.

the probability of picking 5 red marbles: (3/12)⁵, or (1/4)⁵, which is 1/1024.

In case you've picked 4 red marbles

The probability of picking 2 yellow, 1 orange marbles: (5/12)² * (4/12) = 100/1728, or 25/432

The probability of picking 1 yellow, 2 orange marbles: 5/12 * (4/12)² = 80/1728, or 5/108

In case you've picked 5 red marbles

The probability of picking 1 yellow, 1 orange marbles: 5/12 * 4/12 = 20/144, or 5/36

To pick 4 red marbles and still pick at least 1 yellow and 1 orange, we multiply 1/256 with 25/432 and 5/108, so

1/256 * (25/432 + 5/108) = 5/12288

To pick 5 red marbles and still pick at least 1 yellow and 1 orange, we multiply 1/1024 by 5/36, so

1/1024 * 5/36 = 5/36864

To add the probabilities of these two together, we get

5/12288 + 5/36864 = 5/9216

I very much doubt that you would be asked to calculate for picking with replacement in this scenario, but it's good to know how to do it, just in case.

Answer 2

1. There are 3+5+4=12 marbles in the jar in total. The number of ways to choose 3 marbles from the jar is equal to C(12, 3)=(12*11*10)/(3!)=220. The number of ways to choose 3 marbles that contain 2 yellow ones and 1 red and no orange ones is C(5,2)*C(3,1)*C(4,0)=10*3*1=30 since there are 5 yellow, 3 red, and 4 orange marbles. So the probability is 30/220=3/22.
2. Since there are only 3 red marbles, then there is no way that someone can pick 7 marbles with at least 4 being red. So the probability of having at least four red marbles and the remaining being of both yellow and orange color is 0.