Probability of draw with replacement

  • Thread starter Tajeshwar
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In summary: I think perhaps the problem was very badly worded.In summary, the probability of getting the combination of blue and white marbles if the draw is done without replacement is .5.
  • #1
Tajeshwar
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Homework Statement



An urn contains 50 marbles – 40 blue and 10 white. After 50 draws, exactly 40 blue and 10 white are observed.

You are not told whether the draw was done “with replacement” or “without replacement.”

What is the probability that the draw was done with replacement?

Homework Equations

The Attempt at a Solution


[/B]
I am actually not sure where to start on this one. The only thing I can think about is that if the marbles are being replaced the trials would be independent.

Sorry I don't have a better attempt for this one.
 
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  • #2
Tajeshwar said:
Sorry I don't have a better attempt for this one
Not good enough according to the guidelines

Least you can do is provide the relevant equations to calculate the probability of the outcome in the two scenarios. And then attempt something with those :smile: :wink:
 
  • #3
Tajeshwar said:

Homework Statement



An urn contains 50 marbles – 40 blue and 10 white. After 50 draws, exactly 40 blue and 10 white are observed.

You are not told whether the draw was done “with replacement” or “without replacement.”

What is the probability that the draw was done with replacement?

Homework Equations

The Attempt at a Solution


[/B]
I am actually not sure where to start on this one. The only thing I can think about is that if the marbles are being replaced the trials would be independent.

Sorry I don't have a better attempt for this one.

I don't like the question the way it is. In fact, I would claim there is no definite answer. So, I would change it. This may make it even harder to understand or it may give you a key to solving it. My change is:

Your friend has the said jar. They toss a coin. If it's heads, they take 50 balls without replacement; if it's tails, they take 50 balls with replacement. The result is 40 blue and 10 white. What is the probablity the coin was tails?

Hint: probability tree.
 
Last edited:
  • #4
50% if it's an honest coin :biggrin:
Did I break any PF rules stating this ?
 
  • #5
BvU said:
50% if it's an honest coin :biggrin:
Did I break any PF rules stating this ?

I didn't have you down as a probability denier!
 
  • #6
BvU said:
...
Did I break any PF rules stating this ?
Probably not.
 
  • #7
Thank you for your responses.

The first problem I struggled with was understanding P(getting this combination|no replacement) = 1.

After some thought and discussion I figured this out because nothing will be left in the container.

Next I was having trouble with getting some of the values for the bayes theorem formula. I forgot using the law of indifference for choosing one or the other so we can get .5 for one or the other.

Then I used the binomial theorem to get the forward probability part of the problem. 50 C 40 (40/50)^40 (10/50)^10.

Then I used the Bayes theorem with this answer and the .5s from the indifference principle to get P(with replacement|we get this combination 40 blue and 10 white)...

I just got overwhelmed initially because I didn't realize the P(getting combo|without replacement) = 1 and the .5 probability through indifference.
 
  • #8
PeroK said:
I don't like the question the way it is. In fact, I would claim there is no definite answer. So, I would change it. This may make it even harder to understand or it may give you a key to solving it. My change is:

Your friend has the said jar. They toss a coin. If it's heads, they take 50 balls without replacement; if it's tails, they take 50 balls with replacement. The result is 40 blue and 10 white. What is the probablity the coin was tails?

Hint: probability tree.

Yes thank you. This was a big part of what was confusing me. The .5 prob of picking either with replacement or without.
 
  • #9
Tajeshwar said:
Thank you for your responses.

The first problem I struggled with was understanding P(getting this combination|no replacement) = 1.

After some thought and discussion I figured this out because nothing will be left in the container.

Next I was having trouble with getting some of the values for the bayes theorem formula. I forgot using the law of indifference for choosing one or the other so we can get .5 for one or the other.

Then I used the binomial theorem to get the forward probability part of the problem. 50 C 40 (40/50)^40 (10/50)^10.

Then I used the Bayes theorem with this answer and the .5s from the indifference principle to get P(with replacement|we get this combination 40 blue and 10 white)...

I just got overwhelmed initially because I didn't realize the P(getting combo|without replacement) = 1 and the .5 probability through indifference.

I think perhaps the problem was very badly worded.

You wrote "After 50 draws, exactly 40 blue and 10 white are observed." That seems to say that you look into the urn after all the draws have been made, and you see 40 blue and 10 white; that is what the word "after" says to me. Of course, in that interpretation, if you see anything at all in the urn after 50 draws, all those draws must have been made with replacement; so the (conditional) probability sought in the question would be 1.0---no calculation required. However, if it had said "During 50 draws, exactly 40 blue and 10 white were observed", the question would be more meaningful and would require some computations.
 
  • #10
Ray Vickson said:
That seems to say that you look into the urn after all the draws have been made, and you see 40 blue and 10 white; that is what the word "after" says to me
advocate of the dark side. My (naive?) perception is that the 40 / 10 is the outcome (i.e. that what came out of the urn) and you were not allowed to look if the observed marbles were put back or not.

I'm still waiting for @Tajeshwar to come up with the probability of this 40 / 10 outcome for the case that the marbles were put back after observing the colour. Because I do not agree with
Tajeshwar said:
The .5 prob of picking either with replacement or without
In the original story there was no coin toss before the drawing of marbles started.
 
  • #11
Tajeshwar said:
Yes thank you. This was a big part of what was confusing me. The .5 prob of picking either with replacement or without.
On the first draw, what is the probability of picking:
a blue marble?
a white marble?
 

1. What is the concept of "probability of draw with replacement"?

The probability of draw with replacement refers to the likelihood of selecting a specific item from a group of items, and then replacing the selected item back into the group before making another selection. This process continues for a set number of draws, with each draw having the same probability of selecting the desired item.

2. How is the probability of draw with replacement calculated?

The probability of draw with replacement is calculated by dividing the number of desired outcomes by the total number of possible outcomes. For example, if there are 10 red marbles and 20 blue marbles in a bag, the probability of drawing a red marble and replacing it before the next draw would be 10/30 or 1/3.

3. What is the difference between draw with replacement and draw without replacement?

In draw with replacement, the selected item is returned to the group before the next selection, resulting in each draw having the same probability. In draw without replacement, the selected item is not returned to the group, resulting in changing probabilities for each subsequent draw.

4. How does the probability of draw with replacement change as the number of draws increases?

As the number of draws increases, the probability of selecting a specific item with replacement remains constant. This is because each draw is independent of the previous draws, and the probability of selecting the desired item remains the same with each draw.

5. What real-life applications can the concept of probability of draw with replacement be used for?

The concept of probability of draw with replacement is commonly used in various fields such as statistics, genetics, and sampling. In statistics, it is used to calculate the likelihood of a certain outcome in a random experiment. In genetics, it is used to determine the probability of inheriting a certain trait from a parent. In sampling, it is used to estimate the characteristics of a larger population based on a smaller sample size.

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