# Probability and Bernoulli trials

• Lolsauce
In summary, the probability of drawing at least one white ball from a box containing m white and n black balls when k balls are drawn without replacement can be calculated using the formula 1 - (n/m+n)^k, where n is the number of black balls and m is the number of white balls. This is because the probability of drawing at least one white ball is equal to 1 minus the probability of drawing no white balls, which can be calculated using the binomial coefficient (n choose k) divided by the total number of combinations (m+n choose k).
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 * qn-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...

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

Lolsauce
Lolsauce said:

## 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 * qn-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)?

Ray Vickson said:
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?

Buzz Bloom said:
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.

Lolsauce said:
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.

Lolsauce said:
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
Lolsauce said:
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

Last edited by a moderator:
Ray Vickson said:
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.

Ray Vickson said:
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?

Lolsauce said:
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.

Last edited:

## 1. What is probability?

Probability is a measure of the likelihood that a certain event will occur. It is usually expressed as a number between 0 and 1, where 0 represents impossibility and 1 represents certainty.

## 2. What are Bernoulli trials?

Bernoulli trials are a sequence of independent experiments with two possible outcomes: success or failure. Each trial has a fixed probability of success, and the outcome of one trial does not affect the outcome of any other trial.

## 3. How is probability calculated in Bernoulli trials?

The probability of a single success in a Bernoulli trial is calculated using the formula p = n/N, where p is the probability, n is the number of successes, and N is the total number of trials. The probability of no successes (failure) is then 1-p.

## 4. How are Bernoulli trials used in real life?

Bernoulli trials are used in many real-life situations, such as flipping a coin, rolling a dice, or conducting medical experiments. They can also be used to model more complex events, such as stock market fluctuations or election outcomes.

## 5. What is the relationship between Bernoulli trials and the binomial distribution?

The binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials. It is used to calculate the probability of a specific number of successes or failures in a given number of trials, and it is often used in statistical analysis and hypothesis testing.

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