Drawing marked balls from an urn

[Hamiltonian]

Homework Statement

Imagine you have an urn with 200 balls, of which 20 are marked. You draw the balls from the urns one by one, randomly, without putting them back in.
What is the probability of the following events:
(a) The first ball drawn is marked.
(b) The 20th ball drawn is marked.
(c) The 100th ball drawn is marked.
(d) The last ball drawn is marked.

Relevant Equations

$P_{n,k} = \frac{n!}{(n-k)!}$
$\mathcal{P}(E) = \frac{|E|}{|\Omega|}$

part (a) was straightforward $\mathcal{P} = \frac{20}{200} = 0.1$. Instead of directly trying to find the probability of the 20th drawn ball being marked I decided to start with finding the probability of the second ball drawn being marked and then after figuring that out moving to the cases in parts (b)-(d). After the first draw, there are a $199$ balls left hence $|\Omega| = 199$ but there could either be 19 marked balls left in the urn or 20 marked balls left, there is no way of knowing. This is where I am confused and stuck.

 

[Hill]

The probability that there are 19 marked balls left in the urn is the same as the probability that the first ball drawn is marked. The probability that there are 20 marked balls left is 1- the previous probability. Then you can calculate for each case and add.

 

[Hamiltonian]

are u trying to say the probability that the second ball is marked = the probability first ball drawn was not marked + the probability that the first ball drawn was marked?
That doesn't make much sense unless I am misinterpreting what you are saying.

 

[Hill]

No, I am saying that the probability that the second ball is marked = (the probability first ball was not marked) x (the probability second is marked when the first was not) + (the probability first was marked) x (the probability second is marked when the first was).

 

[Hamiltonian]

No, I am saying that the probability that the second ball is marked = (the probability first ball was not marked) x (the probability second is marked when the first was not) + (the probability first was marked) x (the probability second is marked when the first was).

so you are saying,

$$\mathcal{P}(2^{nd}\text{ ball drawn is marked}) = \mathcal{P}(1^{st}\text{ ball not marked})\mathcal{P}(2^{nd}\text{ ball marked when }1^{st}\text{ was not}) + \mathcal{P}(1^{st}\text{ ball is marked})\mathcal{P}(2^{nd}\text{ ball is marked when first also is}) $$
$$=\frac{180}{200}\frac{20}{199} + \frac{20}{200}\frac{19}{199}$$
but I still don't understand why this is correct and how this will extend to calculating the probability of the 20th drawn ball being marked.

 

[Hill]

1. "Why this is correct".
Let's call the event "first ball marked", A, and the event "second ball marked", B.
P(B) = P(B ∧ ¬A) + P(B ∧ A),
P(B ∧ ¬A) = P(¬A) · P(B | ¬A),
P(B ∧ A) = P(A) · P(B | A).
Thus,
P(B) = P(¬A) · P(B | ¬A) + P(A) · P(B | A).

2. "How this will extend ..."
Calculate that number. It will give you a hint.

Alternatively, ask yourself, "after the first ball has been drawn, does the probability of the next ball to be marked go up or down?"

Yet, another way of thinking about it: 10% of all balls are marked; then 10% of any random subset of all balls are marked.

 

[PeroK]

Homework Statement: Imagine you have an urn with 200 balls, of which 20 are marked. You draw the balls from the urns one by one, randomly, without putting them back in.
What is the probability of the following events:
(a) The first ball drawn is marked.
(b) The 20th ball drawn is marked.
(c) The 100th ball drawn is marked.
(d) The last ball drawn is marked.

If you had to bet in this, which ball would you bet has the best chance of being marked? The first, 20th, 100th or 200th?

Or, if you handed each ball to a different person as you drew them, which 20 people out of the 200 would be most likely to get the marked balls?

 

[Hornbein]

Homework Statement: Imagine you have an urn with 200 balls, of which 20 are marked. You draw the balls from the urns one by one, randomly, without putting them back in.
What is the probability of the following events:
(a) The first ball drawn is marked.
(b) The 20th ball drawn is marked.
(c) The 100th ball drawn is marked.
(d) The last ball drawn is marked.
Relevant Equations: $P_{n,k} = \frac{n!}{(n-k)!}$
$\mathcal{P}(E) = \frac{|E|}{|\Omega|}$

part (a) was straightforward $\mathcal{P} = \frac{20}{200} = 0.1$. Instead of directly trying to find the probability of the 20th drawn ball being marked I decided to start with finding the probability of the second ball drawn being marked and then after figuring that out moving to the cases in parts (b)-(d). After the first draw, there are a $199$ balls left hence $|\Omega| = 199$ but there could either be 19 marked balls left in the urn or 20 marked balls left, there is no way of knowing. This is where I am confused and stuck.

Yes, the question is ambiguous. Hence the fine art of figuring out what the teacher wants.

And I'm really not sure! I can think of several reasonable interpretations. In such a case I suppose it is best to say you couldn't figure out what the question was assuming, then state your assumptions and solution. My best guess is that they are assuming you know what all the balls previously drawn were and they want an equation, not a number. But...I'm not sure of that.

 

[PeroK]

I fail to see any ambiguity in the question.

 

[Hornbein]

I fail to see any ambiguity in the question.

I could reasonably respond that the answer to each question is 0.1. This is consistent with these four questions being posed before the drawing of balls begins. That's what I would think that it meant if it didn't run afoul of the "that's too easy" and "not educational" heuristics.

Another interpretation is that the questions are cumulative. (b) is answered with the assumption that the first ball was marked, (c) is answered with the assumption that the first and twentieth were known as marked, and so forth. That could be correct but seems weird to me, that certain balls are known but most of the others not.

If all the results are known then all we can offer is an equation, which falls afoul of the usual practice of calculating a specific number.

If getting the questions out of a textbook usually the author has a certain style the student can come to recognize. One can expect the assumptions to be the same as in other similar questions. For example it's common for "select at random" to mean from a finite uniform distribution without specifically stating such.

In lieu of this, if the question were important enough I'd do the problem all three ways, explain what I was up to, and hope there isn't a fourth interpretation that the author feels is obvious.

 

[scottdave]

so you are saying,

$$\mathcal{P}(2^{nd}\text{ ball drawn is marked}) = \mathcal{P}(1^{st}\text{ ball not marked})\mathcal{P}(2^{nd}\text{ ball marked when }1^{st}\text{ was not}) + \mathcal{P}(1^{st}\text{ ball is marked})\mathcal{P}(2^{nd}\text{ ball is marked when first also is}) $$
$$=\frac{180}{200}\frac{20}{199} + \frac{20}{200}\frac{19}{199}$$
but I still don't understand why this is correct and how this will extend to calculating the probability of the 20th drawn ball being marked.

Did you work out that calculation? How does it compare to the probability of pulling the first ball?

There is an easier way to think about this problem - pull all the balls, placing them in numbered bins without looking at them.

What is the probability that bin # 50 has a marked ball, for example? Or any other chosen position.