# Calculating overall percentage/probability from multiple cat

Probability theory and statistics

### Calculating overall percentage/probability from multiple cat

Greetings,

My apologies if the title is confusing; I don't really know how to explain what I am trying to do or what label this problem would fall under.

Scenario:
We have a magic bag, and inside the magic bag are an unknown/unlimited amount of coins.
There are 100 different types of coins, but we are only interested in the iron, bronze, silver and gold coins, so we have bundled the other 96 types together as "other".

We performed an experiment by taking one coin at a time from the magic bag and recording what type of coin it was; the coins we took from the magic bag were not placed back into the magic bag.
We performed the experiment 1,000,000 times and the results are as follows:

Other coins: 998,859
Iron coins: 596
Bronze coins: 312
Silver coins: 135
Gold coins: 98

First Question:
Is it correct to say the probability to receive each coin from the magic bag is as follows?

Other coins: (998,859/1,000,000)*100 = 99.8859% or 1,000,000/998,859 = 1 in 1.0011423
Iron coins: (596/1,000,000)*100 = 0.0596% or 1,000,000/596 = 1 in 1,678
Bronze coins: (312/1,000,000)*100 = 0.0312% or 1,000,000/312 = 1 in 3,205
Silver coins: (135/1,000,000)*100 = 0.0135% or 1,000,000/135 = 1 in 7,407
Gold coins: (98/1,000,000)*100 = 0.0098% or 1,000,000/98 = 1 in 10,204

Second Question:
If I take only one coin from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).

I have tried to do some calculations, but I don't think I am working it out properly.
- Is the following correct?
((0.0596/100)+(0.0312/100)+(0.0135/100)+(0.0098/100))*100 = 0.1141% (or 100% - 99.8859% = 0.1141%)

- Is the following correct?
((1/1678)+(1/3205)+(1/7407)+(1/10204))*100 = 0.1141%

- Is the following correct?
(((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100 = 0.0285%

Third Question:
If I take 1,400 coins from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).

I really have no idea how to calculate this; all I have managed to do is repeat one of the formulas above and multiply by 1,400.

- Is the following correct?
((((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100)*1400 = 39.9339%

Fourth Question:
What is the name for this type of probability?

I understand that I am probably completely wrong about everything, so thank you very much to anyone willing to provide assistance.
Guest

### Re: Calculating overall percentage/probability from multiple

Guest wrote:

First Question:
Is it correct to say the probability to receive each coin from the magic bag is as follows?

Other coins: (998,859/1,000,000)*100 = 99.8859% or 1,000,000/998,859 = 1 in 1.0011423
Iron coins: (596/1,000,000)*100 = 0.0596% or 1,000,000/596 = 1 in 1,678
Bronze coins: (312/1,000,000)*100 = 0.0312% or 1,000,000/312 = 1 in 3,205
Silver coins: (135/1,000,000)*100 = 0.0135% or 1,000,000/135 = 1 in 7,407
Gold coins: (98/1,000,000)*100 = 0.0098% or 1,000,000/98 = 1 in 10,204

Correct.

Guest wrote:Second Question:
If I take only one coin from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).

I have tried to do some calculations, but I don't think I am working it out properly.
- Is the following correct?
((0.0596/100)+(0.0312/100)+(0.0135/100)+(0.0098/100))*100 = 0.1141% (or 100% - 99.8859% = 0.1141%)

- Is the following correct?
((1/1678)+(1/3205)+(1/7407)+(1/10204))*100 = 0.1141%

- Is the following correct?
(((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100 = 0.0285%

The first two are right. This isn't an average so dividing by 4 is not justified.

Guest wrote:Third Question:
If I take 1,400 coins from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).

I really have no idea how to calculate this; all I have managed to do is repeat one of the formulas above and multiply by 1,400.

- Is the following correct?
((((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100)*1400 = 39.9339%

This isn't correct.
The quickest way to do this one is to work indirectly by first calculating the probabity that all 1400 coins are Other. That's the probability that the 1st coin is an Other, times the probability that the 2nd one is Other, times the probability that the 3rd one is Other ... $= 0.998859^{1400}\approx 0.2022$. That what you don't want so what you do want is $1-0.2022 = 0.7978 = 79.78%$.

Guest wrote:Fourth Question:
What is the name for this type of probability?

Binomial distribution. (Taking an event with only two outcomes and repeating it multiple times.)

phw

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Joined: Sun May 06, 2018 1:15 am
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### Re: Calculating overall percentage/probability from multiple

Hi phw,

I was wondering if it's still possible to calculate the probabilities if we remove some of the data we had initially:

Scenario:
We have a magic bag, and inside the magic bag are an unlimited amount of coins.
We don't know how many types of coins there are, but we are only interested in the iron, bronze, silver and gold coins.
There is a tag attached to the magic bag that states the probability of receiving the following coins:

Iron coins: 0.0596%
Bronze coins: 0.0312%
Silver coins: 0.0135%
Gold coins: 0.0098%

We are not given any other information.

Question 1:
If I take one coin, what is the probability of that one coin being either an iron, bronze, silver or gold coin?
I mean to say that I am putting the iron, bronze, silver and gold coins into a group of preferred coins, and receiving any of them would be a success.

I think the following would be correct:
- ((0.0596/100)+(0.0312/100)+(0.0135/100)+(0.0098/100))*100 = 0.1141%

Question 2:
If I take 1,400 coins, what is the probability that among those 1,400 coins, at least one of them is an iron, bronze, silver or gold coin?
I mean to say that I am putting the iron, bronze, silver and gold coins into a group of preferred coins, and receiving at least one of any of them would be a success.

Is the following correct?
- (1-(((1-(0.0596/100))*(1-(0.0312/100))*(1-(0.0135/100))*(1-(0.0098/100)))^1400))*100 = 79.7646%

Thanks!
Guest