Probability question---how to create formulas.?

Question

1)You have a container.
It contains ten marbles: 4 red, 3 blue, 2 white, 1 black
If you remove one stone and put it in your pocket without looking at it, what is the probability that the next marble you pull out of the container will be red?

What would the probability be for it to be black?

2) Scenario is same except you have removed six marbles from container and put them in your pocket (without looking at them). What is the probability that the next one from the container is red?

What is the probability that it would be blue?

Ultimately I need a formulaic way to compute this out for a larger number of items with various quantities of each. So it is more of the computational way to do this than the requirement of an actual numeric answer. Thanks.

Response:

So you are saying that the probability does not change since we cannot varify the depletion of any of the contents that are being moved from one container to another.

It is hard for me to believe that the probability of black will remain 1/10 regardless of the number of items in container...even at one item left.

Answer 1

Without any information about the first draw, all marbles are equally likely to be drawn second.
Therefore

Prob(second draw is red) = 4/10 = 2/5.

The longer way is
P(R then R) + P(non R then R) = (4/10)*(3/9) + (6/10)*(4/9)
= 2/15 + 4/15
= 6/15
= 2/5
See the advantage of logic over formulas!

By the same logical principle,
P(second is black) = 1/10. You can check it the long way if you like.

2) Same reasoning tells me P(sixth draw is red) = 2/5.
P(blue) = 3/10

A longer approach to this would be to say
Number of ways of filling six positions with marbles chosen from these ten is
10P6 = 10*9*8*7*6*5

Number of ways of filling the sixth position with one of the four red marbles and then filling the other five with marbles chosen from the remaining nine is
4* (9P5) = 4*9*8*7*6*5

So P(sixth is red) = (4*(9P5))/(10P6)
which is a long way of getting 2/5, as before.

PS (After reading your response). Yes, that's what I'm saying. Even if you find that hard to believe, try computing it yourself:

P(second is black) = P(first is not black)*P(second is black if the first is not black).

The thing is, if we don't know anything about the prior draws, we're just saying: "We're going to draw some marbles. What's the probability the second (or third, or sixth) one will be red (or black, or blue)?"
email me or [e-mail] if you want to discuss it further.

Answer 2

The probability does not change, it is 4/10.

Before you pull out a marble, the probability that the marble you pull will be red is 4/10.

Then you pulled a marble, but you did not look at it. If you had looked at it, your probabilities would change, because you had new information. But you did not look at it, so you have no new information.

Let's consider the problem this way. Suppose you pull one marble from the container, and do not look at it, and put it in your pocket. Now, you pull a second marble, and do not look at it, and put it in your pocket. In the same fashion, you pull nine marbles, and do not look at them, and put them in your pocket. What is the probability that the one marble remaining in the container is red? It's 4/10.

You are thinking that you have changed the problem by puttint the marble in your pocket, but you have not. You still have ten marbles, you have changed the location of one, but have not changed your knowledge about the percentages in the container or in your pocket. The probability that the one in your pocket is red is 4/10, and the probability that the next one you pull from the container is also 4/10.

Answer 3

Hy is correct.
This is called un-conditional probability.

You see, if you don't know what color marble you put in your pocket, there are 4/10 chance that you removed a red, 3/10 chance that you removed a blue, 2/10 white, 1/10 black. Therefore the entire chances of all 4 color got scaled down. At the same time, the total marble also got scaled down.

If 10 people picked a marble (but can't see the color yet), and you are the last person, the chance is the same as first person. Because this last marble left in the container (even only 1 marble left), it has 4/10 chance of being red, 3/10 chance of being blue...... You can think of it as the the chances, not as counting game.

Answer 4

You're actually asking conditional probability questions.

For example, if that first marble was black then there wouldn't be any black to pull:

P( 2nd pull black | first pull black ) = 0

Same thing on the scenario 2. You're posing conditional probability questions the answers to which depend upon the first action.