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Explain binomial probability to a mathlet

Anonymous Fri 03 Jan 15:08:20 2020 No.11274182 View Reply Original Report

Quoted By: >>11274203 >>11274260 >>11275387

College algebra student, what is binomial probability in mathlet terms

Anonymous

Anonymous Fri 03 Jan 2020 15:17:48 No.11274203 Report

Quoted By:

>>11274182
just a systematic way of keeping tabs of the hits and misses

Anonymous

Anonymous Fri 03 Jan 2020 15:45:25 No.11274260 Report

Quoted By:

>>11274182

P(x) is the probability that you get some set of x successes after n trials.

(n choose x) is the number of possible ways to choose the x trials that succeed from the n trials that fail. For example, if I'm flipping a coin 5 times and I want to get 4 successes, I could get any of the following outcomes:

1 - 01111
2 - 10111
3 - 11011
4 - 11101
5 - 11110

Each coin flip is an independent event, so the probability of each of the sequences is:

1 - q*p*p*p*p
2 - p*q*p*p*p
3 - p*p*q*p*p
4 - p*p*p*q*p
5 - p*p*p*p*q

Notice that all of these products are just equal to $p^x(1-p)^{n-x}$ .

In other words, what you're doing is saying:

"Probability that we get a successful event" = "number of possible successful events"*"probability of a successful event"

Anonymous

Anonymous Fri 03 Jan 2020 18:03:57 No.11274514 Report

Quoted By:

Let's say there are a number of combinations, each having an equal probability of occurring. However, many of those combinations sum to the same final result. So for example, two coin flips, with head being worth "1" and tails being worth "0"

HH = 2
HT = 1
TH = 1
TT = 0

So you see, even though each of these combinations is equally likely, you are more likely to get a "1" since there's more combinations that lead to this result.

Let's say that you have a certain probability of stepping left "q", and a certain probability of stepping right "p" (notice that p = 1-q). After 5 steps, just 1 of the combinations looks like this
p q q p p
That is three steps right, and two steps left. However, there are more combinations that result in the same distance traveled:
q q p p p also gets you there. And p p q p q.....etc.
The probability of each singular instance occurring is just p^3*q^2. However, there are many different combinations which need to be added together.
Summing up all of the combinations, we need to know how many different ways 3 p's and 2 q's can be placed in order.

There is a way to prove that this is n!/(n-x)!x! where n is the number of steps and x is the number of steps to the right.

Anonymous

Anonymous Sat 04 Jan 2020 01:03:04 No.11275387 Report

Quoted By:

>>11274182
If there is a probability of 0.25 everytime you cum in the pussy that the bitch it's going to get pregnant, then if you cum 4 times it's like 4*0.25 = 1
But you know this is not true cuz Jamal TOPPED your tranny ass more than 4 times.
Then WTF it's happening you ask yourself
Well there are 2 main problem about this:
>1 Tranny doesn't get pregnant
>2 You have to use a binomial distribution

We cumulate the probability and we get 0.683

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