# How to handle probabilities of the number of trials in a Binomial distribution

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• benorin
In summary: The probability that neither Guns nor Butter has occurred after we do the second trial is .85.The probability that neither Guns nor Butter has occurred exactly after we do the third trial is .15.The weighted average of these is .857.
benorin
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TL;DR Summary
Suppose our process has a 85% chance of 2 trials and a 15% chance of 3 trials, and the rest is straightforward binomial distribution, do I take the weighted average?
Suppose our process has a 85% chance of 2 trials and a 15% chance of 3 trials, and the rest is straightforward binomial distribution, do I take the weighted average of the binomial distribution at 2 and 3 trials? This is for a game so, yeah thanks.

benorin said:
Summary: Suppose our process has a 85% chance of 2 trials and a 15% chance of 3 trials, and the rest is straightforward binomial distribution, do I take the weighted average?

Suppose our process has a 85% chance of 2 trials and a 15% chance of 3 trials, and the rest is straightforward binomial distribution, do I take the weighted average of the binomial distribution at 2 and 3 trials? This is for a game so, yeah thanks.

What are you trying to calculate?

What do you mean by "taking"? What are you taking? And what are the parameters of your binomial? What is " The rest"? Please elaborate a bit so we can understand.

Stephen Tashi
Well the purchase of a pack of materials in a game has say a 85% chance of 2 pulls (trials) and a 15% chance of 3 pulls of a selection of materials each of which has a fixed % chance. Say the sum of desirable chances is X (in decimal form), I wish to calculate the probability of getting one or more of the set of desirable materials. I can easily calculate the chance for 2 and 3 trials, but do I just take a weighted average of these?

benorin said:
Well the purchase of a pack of materials in a game has say a 85% chance of 2 pulls (trials) and a 15% chance of 3 pulls of a selection of materials each of which has a fixed % chance. Say the sum of such chances is X (in decimal form), I wish to calculate the probability of getting one or more of the set of desirable materials. I can easily calculate the chance for 2 and 3 trials, but do I just take a weighted average of these?
Yes.

FactChecker and benorin
PeroK said:
Yes.

(!) What is your interpretation of the question?

Stephen Tashi said:
(!) What is your interpretation of the question?
Good question! I assumed we have a) probability of something (trials = 2); b) probability of that same something (trials = 3). And we're looking for the overall probability of that something, given the probability of a) and b).

Yes. Technically this is called a mixture of two populations and you can treat the probability of some outcome ##A## as conditioned on what population you pulled from.

##P(A) = P(A | B_1) P(B_1) + P(A | B_2) P(B_2)## where ##B_1## and ##B_2## are the events that you pulled from population 1 or population 2, respectively.

FactChecker
Let's guess what the question is.

benorin said:
Well the purchase of a pack of materials in a game
I'll imagine the pack of materials is guns and butter.

has say a 85% chance of 2 pulls (trials)
Can you "pull" both guns an butter on the same "pull" or can you only get one of them per pull?
There's an .85 probability that both guns and butter will have been pulled after the completion of the second pull ?
and a 15% chance of 3 pulls of a selection of materials each of which has a fixed % chance.
Assume that on one pull there is a probability of G to pulls guns and a probablity of B to pull butter, and at most one of Guns and Butter can be pulled on a given pull. Is this saying there is a .15 probability that both guns and butter will have been pulled exactly after the 3rd pull? - i.e. not before then, not on the second pull.

Say the sum of desirable chances is X (in decimal form),
Assume P+B = X.

I wish to calculate the probability of getting one or more of the set of desirable materials.
You want to calculate the probability of getting at least one pair of Guns or Butter after N pulls for N = 4,5,6,...(?)

I can easily calculate the chance for 2 and 3 trials, but do I just take a weighted average of these?
What would a "weighted average" be? ##\frac{ (3)(.85) + (2)(.15)}{ (3 + 2)} ## ? What would you use it to calculate?

The title mentions "binomial distribution" I'll try this version:

On each trial there are 3 mutually exclusive outcomes, Guns, Butter, and Neither Guns nor Butter. The event Guns occurs with probability G. The event Butter occurs with probability B. The event Neither guns nor butter occurs with probability 1 - P - B.

The trials are independent. We are given the following:
The probability that both Guns and Butter have occurred after we do the second trial is .85.
The probability that both Guns and Butter have occurred exactly after we do the third trial is .15 (i.e. Both guns and butter have occurred after the third trial, but have not both occurred just after the second trial)
The value of P+B = X is known.

Question: Find the probability that both Guns and Butter have occurred at least once after N trials for N = 4,5,6,...

It would be an different problem if both Guns and Butter have a chance of being produced by the same pull.
If you can calculate G_k = the probability that at least one occurence of Guns occurs by the the end of the kth trial and B_k = the probability that at least one occurence of Butter occurs by the kth trial, and these events are independent, then the product (G_k)(B_k) is the probability that both events occur at least once by the end of the kth trial.

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My take is that you want your favorite , say, baseball card. Cards come in different packs. Your first pack has your card with probability .15 and the second one has .85 probability of containing it. You then want to know the probability of obtaining your card if you buy both. This is ##P(C|A \cup B)##. Is this the layout?

WWGD said:
My take is that you want your favorite , say, baseball card. Cards come in different packs. Your first pack has your card with probability .15 and the second one has .85 probability of containing it. You then want to know the probability of obtaining your card if you buy both. This is ##P(C|A \cup B)##. Is this the layout?

Some advisors see "pack of materials" as a pack of the same kind of thing. I'm visualizing it as consisting of different types of things. The statement of the problem isn't clear, so you might be right.

WWGD
Stephen Tashi said:
Some advisors see "pack of materials" as a pack of the same kind of thing. I'm visualizing it as consisting of different types of things. The statement of the problem isn't clear, so you might be right.
Agreed, it is ambiguous; either of us may be right. It seems OP is no longer around to clarify.

No. a pull is a draw from the pool of items with given distribution. We have an 85% chance of two draws and a 15% chance of three draws.

benorin said:
No. a pull is a draw from the pool of items with given distribution. We have an 85% chance of two draws and a 15% chance of three draws.

That description is unclear. It's also unclear what statement you are saying "No" to. I suggest you give a specific description of the problem - or describe an example of a similar but simpler problem.

The distribution would be something like {A-60%, B-30%, C-10%}. If purchasing a pack gives an 85% chance of two draws from the distribution and a 15% chance of three draws from the distribution, determine the probability that a pack contains at least 1 of B or C? Here's a handy binomial calculator to ease your labors.

Let B(n,p,x) denote binomial probabilities for n trials with p probability of success and at least x successes. Then would the prior post be answered by P(at least one B or C)= 0.85*B(2,0.4,1)+0.15*B(3,0.4,1) ?

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*bump*

Yes.

benorin

## 1. What is a Binomial distribution?

A Binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent trials, where each trial has the same probability of success. It is often used to model situations where there are only two possible outcomes for each trial, such as heads or tails in a coin toss.

## 2. How do you calculate the probability of a specific number of successes in a Binomial distribution?

The probability of a specific number of successes in a Binomial distribution can be calculated using the formula P(x) = nCx * p^x * (1-p)^(n-x), where n is the number of trials, x is the number of successes, and p is the probability of success in each trial. nCx represents the number of combinations of x successes out of n trials.

## 3. What is the expected value of a Binomial distribution?

The expected value of a Binomial distribution is equal to n * p, where n is the number of trials and p is the probability of success in each trial. This represents the average number of successes that can be expected in a given number of trials.

## 4. How do you handle probabilities for a large number of trials in a Binomial distribution?

For a large number of trials in a Binomial distribution, it is often more convenient to use the Normal approximation. This involves calculating the mean and standard deviation of the Binomial distribution and then using the Normal distribution to estimate probabilities.

## 5. Can the Binomial distribution be used for continuous data?

No, the Binomial distribution is only applicable for discrete data where there are a fixed number of trials and only two possible outcomes for each trial. For continuous data, other distributions such as the Normal distribution may be more appropriate.

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