MHB Calculating overall percentage/probability from multiple categories?

[Spud]

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 10,204Second 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%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%I understand that I am probably completely wrong about everything, so thank you very much to anyone willing to provide assistance.

 

[Chris L T521]

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

The types of probabilities that must be computed here are empirical probabilities. The first values you have listed would be the more appropriate way of stating the answers.

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%

Since these are mutually exclusive events, you can use this form of the addition rule:

$P(I\cup B \cup S\cup G) = P(I)+P(B)+P(S)+P(G)$.

In this case, we find that

$P(I\cup B \cup S\cup G) = 0.0596\% + 0.0312\% + 0.0135\% + 0.0098\% = 0.1141\%$

Alternatively, you could apply the complement rule to get $P(I\cup B\cup S\cup G) = 1 - P(O) = 100\% - 99.8859\% = 0.1141\%$.

Thus, the first way you did the problem would be the best way to do it.

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%

Are we assuming here that we're finding the probability that exactly one of the coins from the 1400 is iron, bronze, silver, or gold? If so, then we can use the Hypergeometric distribution to answer this question because the coins are being selected without replacement. (On the other hand, if the coins were being selected with replacement, we could have used the binomial distribution instead.)

From the wiki link, if we're given a finite population of size $N$, and $K$ is the number of objects with the desired trait, then the probability of picking $k$ successes in $n$ trials is given by

$P(X=k) = \dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$.

In this problem, $N=1000000$, $n=1400$, $K=1141$, and $k=1$.

Thus, $P(X=1) = \dfrac{\binom{1141}{1}\binom{1000000-1141}{1400-1}}{\binom{1000000}{1400}}\approx 32.3515\%$

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

You seemed to be on the right track about the first two. The last part was pretty tricky since the wording of the problem was vague (in my opinion). If my interpretation for the last part is incorrect, please let me know and we'll try to figure things out from there! \o/

I hope this helps!

 

[Spud]

Thanks Chris L T521! Very detailed and helpful.

I am happy that my level of understanding wasn't too far off track :)

So in regards to the last question with the 1,400 coins; I don't think I was very clear. I will try to clarify in a moment, but firstly I want to adjust the scenario and perhaps that will make things more clear.

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.

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.I hope this is more clear; my apologies if it's not.
Thank you kindly.

 

[HOI]

With an unlimited number of coins, whether we return coins or not doesn't matter so we use the binomial distribution.