Probability and Bernoulli trials

[Lolsauce]

Homework Statement

A box contains m white and n black balls. Suppose k balls are drawn. Find the probability of drawing at least one white ball.

Homework Equations

Probability of one success = P({1 successful trial}) = n * p * q^n-1
p = probability
where q = 1-p
Fundamental theorem of Bernoulli trials (or k successes):

The Attempt at a Solution

My sample size is m+n. So the probability of white is:

P(W) = m/(m+n)

There are k balls drawn. I did not know if we want the equation for 1 successful trial or k successes, as there is a probability of getting a white ball more than once. I went with the first equation as the keywords "at least one white ball". Using the first equation I get:

p = P(W)
q = 1 -P(W)
P({1 success}) = n (m / (m+n))ⁿ(1 - (m / (m+n))^n-1

I looked at the solution manual and have no idea how they got the following...

 

[Buzz Bloom]

Hi Lolsause:

I am guessing that your confusion may be that your are considering the selection of a ball on trial k to have the same probability that the ball is white for all trials. That would be the case if the selected ball is returned to the box after each trial. But the statement, "Suppose k balls are drawn," means that balls are not returned to box after each trial.

I hope this helps.

Regards,
Buzz

 

[Ray Vickson]

Homework Statement

A box contains m white and n black balls. Suppose k balls are drawn. Find the probability of drawing at least one white ball.

Homework Equations

Probability of one success = P({1 successful trial}) = n * p * q^n-1
p = probability
where q = 1-p
Fundamental theorem of Bernoulli trials (or k successes):

The Attempt at a Solution

My sample size is m+n. So the probability of white is:

P(W) = m/(m+n)

There are k balls drawn. I did not know if we want the equation for 1 successful trial or k successes, as there is a probability of getting a white ball more than once. I went with the first equation as the keywords "at least one white ball". Using the first equation I get:

p = P(W)
q = 1 -P(W)
P({1 success}) = n (m / (m+n))ⁿ(1 - (m / (m+n))^n-1

I looked at the solution manual and have no idea how they got the following...

The probability formula you wrote in (2) is for the Binomial distribution, which would hold only if you put back each ball after drawing it out. If (as is probably intended) the balls are drawn without replacement, then you do not have a Bernoulli process. You need to use a different distribution.

In your case you do not want the probability of exactly a certain number of white balls, but instead want P(>= 1 white). Do you see why this equals 1-P(0 white)? Do you see how to compute P(0 white)?

 

[Lolsauce]

The probability formula you wrote in (2) is for the Binomial distribution, which would hold only if you put back each ball after drawing it out. If (as is probably intended) the balls are drawn without replacement, then you do not have a Bernoulli process. You need to use a different distribution.

In your case you do not want the probability of exactly a certain number of white balls, but instead want P(>= 1 white). Do you see why this equals 1-P(0 white)? Do you see how to compute P(0 white)?

So a Bernoulli trial assumes the balls are put back as it's the trial of the whole system per k trials?

I see why the equals 1-P(0 white balls). We are essentially finding the compliment of getting a black ball correct?

I'm not exactly sure how they computed P(0 white balls). I guess it's because I'm not familiar with the choose/combination notation.

From what I see the probability of P(0 white balls) is suppose to be...
(n choose k)/ (m+n choose k)

Meaning we want the probability of k subsets in n (total k black) WITHIN k subsets in our total sample size m + n (total black and white).

OHHH I think I get it. Is my assumption correct?

 

[Lolsauce]

Hi Lolsause:

I am guessing that your confusion may be that your are considering the selection of a ball on trial k to have the same probability that the ball is white for all trials. That would be the case if the selected ball is returned to the box after each trial. But the statement, "Suppose k balls are drawn," means that balls are not returned to box after each trial.

I hope this helps.

Regards,
Buzz

Ok, I can see that now with you and the other users reply. This is not a Bernoulli trial, as the balls are not returned. A Bernoulli experiment is in the same sample space.

 

[Ray Vickson]

So a Bernoulli trial assumes the balls are put back as it's the trial of the whole system per k trials?

I see why the equals 1-P(0 white balls). We are essentially finding the compliment of getting a black ball correct?

I'm not exactly sure how they computed P(0 white balls). I guess it's because I'm not familiar with the choose/combination notation.

From what I see the probability of P(0 white balls) is suppose to be...
(n choose k)/ (m+n choose k)

Meaning we want the probability of k subsets in n (total k black) WITHIN k subsets in our total sample size m + n (total black and white).

OHHH I think I get it. Is my assumption correct?

Never mind the binomial coefficients for now; the first step is to understand what is happening.

Let's do a simple example, with 5 white and 5 black balls. We choose two balls at random, without replacement, and want to know P(0 white) = P(2 black). Let the events be B1={first is black} and B2 = {2nd is black}. P(B1) = 5/10. Now what? After drawing a black there are 9 balls left, of which 4 are black, so P(B2|B1) = 4/9. Thus, P(B1 & B2) = P(B1) * P(B2|B1) = (5/10)(4/9).

OK: over to you.

 

[Buzz Bloom]

Is my assumption correct?

Aj:

I confess I do not understand the notation you are using. I also am not sure I understand exactly what your "assumption" is.

Here is a suggestion.
Let D(i) be the proposition: The ball drawn for trial i is black.
Then: Prob{all trials draw a black ball} = Prob{D(1)} × Prob{D(2)} × Prob{D(3)} × . . . Prob{D(N)}

Wikipedia
https://en.wikipedia.org/wiki/Bernoulli_trial
says: "Bernoulli trial (or binomial trial) is a random https://www.physicsforums.com/javascript:void(0) with exactly two possible https://www.physicsforums.com/javascript:void(0) , "success" and "failure", in which the probability of success is the same every time the experiment is conducted."
From this definition is appears that you are right that

a Bernoulli trial assumes the balls are put back as it's the trial of the whole system per k trials

Why do you think this problem is about Bernoulli trials? The solution you quote seem to assume that the balls are NOT replaced.

Regards,
Buzz

 

[Lolsauce]

Never mind the binomial coefficients for now; the first step is to understand what is happening.

Let's do a simple example, with 5 white and 5 black balls. We choose two balls at random, without replacement, and want to know P(0 white) = P(2 black). Let the events be B1={first is black} and B2 = {2nd is black}. P(B1) = 5/10. Now what? After drawing a black there are 9 balls left, of which 4 are black, so P(B2|B1) = 4/9. Thus, P(B1 & B2) = P(B1) * P(B2|B1) = (5/10)(4/9).

OK: over to you.

Okay, I understand this.

 

[Lolsauce]

Never mind the binomial coefficients for now; the first step is to understand what is happening.

Let's do a simple example, with 5 white and 5 black balls. We choose two balls at random, without replacement, and want to know P(0 white) = P(2 black). Let the events be B1={first is black} and B2 = {2nd is black}. P(B1) = 5/10. Now what? After drawing a black there are 9 balls left, of which 4 are black, so P(B2|B1) = 4/9. Thus, P(B1 & B2) = P(B1) * P(B2|B1) = (5/10)(4/9).

OK: over to you.

So by applying this to my problem.

P(0 white) = (n/(m+n)) * (n-2)/(m+n) * ... (n-k+1)/(m+n) = n!/(n-k)!

Does this look correct?

 

[Ray Vickson]

So by applying this to my problem.

P(0 white) = (n/(m+n)) * (n-2)/(m+n) * ... (n-k+1)/(m+n) = n!/(n-k)!

Does this look correct?

No. You already wrote down the correct formula from the solutions manual, and it does not match up with what you have written just now.