Probability of 11 to be in 10 draws without return from 20

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valnoir

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Hello,

Please let me know if the solution I thought of is the right one:


From a bag containing 20 balls numbered from 1 to 20 will be drawn 10 balls. What is the probability that the ball with number 11 will be among the 10 balls drawn without return. So on the second draw, there are 19 balls in the bag.

Notations:

P (1) = the probability that ball number 11 will be drawn in the 1st draw

P (2) = the probability that ball number 11 will be drawn in the 2nd draw

P (3) = the probability that ball number 11 will be drawn in the 3rd draw

P (4) = the probability that ball number 11 will be drawn in the 4th draw

...

P (10) = the probability that ball number 11 will be drawn in the 10th draw

P (A) = the probability that ball number 11 will be drawn in any of the 10 draws

Solution:

P (A) = P (1) + P (2) + P (3) + P (4) + ... + P (10)

P (A) = 1/20 + 1/19 + 1/18 + 1/17 + ... + 1/11

P (A) = 0.668771403175428


Thank you!
 
valnoir said:
Hello,

Please let me know if the solution I thought of is the right one:


From a bag containing 20 balls numbered from 1 to 20 will be drawn 10 balls. What is the probability that the ball with number 11 will be among the 10 balls drawn without return. So on the second draw, there are 19 balls in the bag.

Notations:

P (1) = the probability that ball number 11 will be drawn in the 1st draw

P (2) = the probability that ball number 11 will be drawn in the 2nd draw

P (3) = the probability that ball number 11 will be drawn in the 3rd draw

P (4) = the probability that ball number 11 will be drawn in the 4th draw

...

P (10) = the probability that ball number 11 will be drawn in the 10th draw

P (A) = the probability that ball number 11 will be drawn in any of the 10 draws

Solution:

P (A) = P (1) + P (2) + P (3) + P (4) + ... + P (10)

P (A) = 1/20 + 1/19 + 1/18 + 1/17 + ... + 1/11

P (A) = 0.668771403175428


Thank you!
Click to expand...
No, that's incorrect.

If you do it this way, you need to calculate P(2) as the probability that 11 is not drawn first, AND is drawn second. This is 19/20 * 1/19. Do you see why? Then do the rest. (This is needed because if it had been already drawn, then its probability would be 0, not 1/19!)

If I were you, I would try calculating this a couple different ways, in order to check and also learn multiple ways. Combinations and permutations would be other good options.

Or you might just consider the situation after the 10 balls are chosen, so you have two equal piles, one of chosen balls and one of remaining balls. What is the probability that the 11 is in the former?
 
valnoir said:
From a bag containing 20 balls numbered from 1 to 20 will be drawn 10 balls. What is the probability that the ball with number 11 will be among the 10 balls drawn without return. So on the second draw, there are 19 balls in the bag.
Click to expand...
This is a confused & confusing post..
If it is without return then how is it that nineteen are left?
Is it that if the eleven ball i among the first ten drawn then the other nine ate returned?
What happens if the eleven ball is not in the first ten drawn?

[imath][/imath][imath][/imath]
 
Last edited by a moderator:
pka said:
If it is without return then how is it that nineteen are left?
Click to expand...
valnoir said:
From a bag containing 20 balls numbered from 1 to 20 will be drawn 10 balls. What is the probability that the ball with number 11 will be among the 10 balls drawn without return. So on the second draw, there are 19 balls in the bag.
Click to expand...

I interpret "without return" as "without replacement", either mistranslated or just a little oddly stated, and misplaced (as one using a second language might). Balls drawn are not returned to the bag. That makes the question sensible:

From a bag containing 20 balls numbered from 1 to 20, 10 balls will be drawn, without replacement. For example, on the second draw, there are 19 balls in the bag.​
​
What is the probability that the ball with number 11 will be among the 10 balls drawn?​
 
It appears that you have a problem not written in English and badly translated.

Here is what I guess the problem is.

There is a bag containing 20 indistinguishable balls except each is uniquely numbered from one to twenty. If ten balls are drawn at random from the bag without any replacement, what is the probability that one of those balls is numbered 11?

Is that the problem?

One way you might try is:

[math]\dfrac{1}{20} + \dfrac{1}{19} * \dfrac{19}{20} + … \dfrac{1}{10} * \dfrac{10}{20}[/math]
What’s that equal? What is the logic? Is that logic sound? Why?

The other way to do it is

[math]1 - \left ( \dbinom{19}{10} \div \dbinom{20}{10} \right ).[/math]
What is the logic? Is that logic sound? Why?

Do the answers from the two methods agree?

If both methods agree and seem logical, what do you think the correct answer is?
 
pka said:
This is a confused & confusing post..
Click to expand...
So are several of your posts, pka. The large number of typographical errors in your posts often make your statements difficult to understand. Like this one:
Is it that if the eleven ball i among the first ten drawn then the other nine ate returned?
Click to expand...

Kindly use the preview button to proofread and correct your posts before submitting them. Thank you!

?

[imath]\;[/imath]
 
There are 20 balls. You pick one at a time until you put 10 in your left pocket. The remaining 10 go in your right pocket. What is the probability that ball 11 is in your left pocket?
 
Steven G said:
There are 20 balls. You pick one at a time until you put 10 in your left pocket. The remaining 10 go in your right pocket. What is the probability that ball 11 is in your left pocket?
Click to expand...
Yes, this is the problem.
 
So far I understand that I have to take in consideration the probability that the ball is not picked first time, 2nd time,...,9th time.
As far as theory goes I know this :
[math]P(B|A)=\frac{P(A\cap B)}{P(B)}\\[/math]If the events are independent then
[math]P(A\cap B)=P(A)*P(B)[/math]So we would obtain
[math]P(B|A)=\frac{P(A)*P(B)}{P(B)}=P(A)[/math]So in my case, I would obtain
[math]P(A)=P(1)+P(\overline{1}|2)+P(\overline{2}|3)+...+P(\overline{19}|20)[/math][math]P(A)=P(1)+P(\overline{1})+P(\overline{2})+...+P(\overline{19})[/math][math]P(A)=\frac{1}{20}+\frac{19}{20}+\frac{18}{19}+...+\frac{11}{12}[/math][math]P(A)=8.47213768773366[/math]Which can't be true because
[math]0 \leq P(A) \leq 1[/math]So I guess that the events are dependent and the correct answer would be
JeffM said:
One way you might try is:


[math]P(A)=\frac{1}{20} + \dfrac{1}{19} * \dfrac{19}{20} + … \dfrac{1}{10} * \dfrac{10}{20}[/math]
Click to expand...
Which is
[math]P(A)=\frac{1}{20} + \dfrac{1}{20} + … +\dfrac{1}{20}[/math][math]P(A)=\frac{10}{20}=0.5[/math]
Am I right?
 
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valnoir said:
So far I understand that I have to take in consideration the probability that the ball is not picked first time, 2nd time,...,9th time.
As far as theory goes I know this :
[math]P(B|A)=\frac{P(A\cap B)}{P(B)}\\[/math]If the events are independent then
[math]P(A\cap B)=P(A)*P(B)[/math]So we would obtain
[math]P(B|A)=\frac{P(A)*P(B)}{P(B)}=P(A)[/math]So in my case, I would obtain
[math]P(A)=P(1)+P(\overline{1}|2)+P(\overline{2}|3)+...+P(\overline{19}|20)[/math][math]P(A)=P(1)+P(\overline{1})+P(\overline{2})+...+P(\overline{19})[/math][math]P(A)=\frac{1}{20}+\frac{19}{20}+\frac{18}{19}+...+\frac{11}{12}[/math][math]P(A)=8.47213768773366[/math]Which can't be true because
[math]0 \leq P(A) \leq 1[/math]So I guess that the events are dependent and the correct answer would be

Which is
[math]P(A)=\frac{1}{20} + \dfrac{1}{20} + … +\dfrac{1}{20}[/math][math]P(A)=\frac{10}{20}=0.5[/math]
Am I right?
Click to expand...
Yes, you are right. But do you understand the logic behind the method?
 
JeffM said:
Yes, you are right. But do you understand the logic behind the method?
Click to expand...
Yes, I understand that the event that the ball to be picked 2nd depends on the fact that the ball is not picked 1st, so this is the reason that I need to account for the probability of not 1 [math](\overline{1})[/math].
Thank you!
 
Steven G said:
There are 20 balls. You pick one at a time until you put 10 in your left pocket. The remaining 10 go in your right pocket. What is the probability that ball 11 is in your left pocket?
Click to expand...
valnoir said:
Yes, this is the problem.
Click to expand...
You have 20 identical balls. You pick 10 balls to put in your left pocket and place the other 10 balls in your right pocket.
The probability that ball 11 is in your right pocket is the same as the probability that the ball is in your right left pocket. The answer is 1/2.
 
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Steven G said:
You have 20 identical balls. You pick 10 balls to put in your left pocket and place the other 10 balls in your right pocket.
The probability that ball 11 is in your right pocket is the same as the probability that the ball is in your right pocket. The answer is 1/2.
Click to expand...
Proofread please.

Otherwise, good explanation. And when two methods give the same answer, there's a good chance it's correct.
 
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