Question

Stats question: I just wanna verify if I got the correct answer.
"Three coins, consisting two 10$, three 5$, and five 1$, are to be selected randomly. If we let X be the total amount of the coin. What is the expected mean value of X?"
My answer is 12, can somebody just verify it. :)

Answer 1

what did you do to get 12?

Answer 2

I made table.

Answer 3

I've been thinking about this. There are 10 coins in total and each "try" consists of randomly selecting 3. Basically, we are selecting 3 objects out of a set of ten. If we use the formula n!/(n-k)!*k!
we can see there are
10! / (7! * 3!) = 10*9*8 / 6 =
120 ways in which they can be chosen.
Let's suppose we wanted to calculate the odds of getting a $3 total.
There are 10 ways in which 3 objects can be chosen from 5.
So probability of choosing $3 = 10/120 = .083333333

Answer 4

I actually made a table for this.
x f(x)
3 (getting three 1$ coin) then i solved for f(x)
7 (getting one 5$ coin and two 1$ coins)
11 (getting two 5$ coin and 1$ coin)
12 (getting one 10$ coin and 2$ coin)
15 (getting three 5$ coin)
16 (getting one coin per 1$ 5$ and 10$)
21 (getting two for 10$ and one for 1$)
25 (getting two for 10$ and one for 5$)

Answer 5

Since \(E(X)=\displaystyle\sum_{\text{all }x\text{'s}}x\cdot f(x)\), I think the first step would be to find the range of \(X\), then determine the probability each combination that gives you a particular \(X\).

For example, you can have 2 $10 coins and 1 $5 coin, so \(X=25\). The probability of getting this combo of coins would be \(\dfrac{\binom22\binom21}{\binom{10}3}\).

Answer 6

Ah, looks like that's what you've done so far.

Answer 7

That fraction should have a \(\binom31\), not \(\binom21\), sorry.

Answer 8

yttrium
I see you have all possible totals listed but (as you may know) their probabilities are not equally likely.

Answer 9

For example (as I posted previously) the probability of choosing a $3 total is .08333333

Answer 10

Yeah. I know they don't have equal possibilities. So what I did is that, say for x= 3
\[\frac{ 5C3 }{ 10C3 }\]

Answer 11

The result will then be equal to f(3). Any reaction to that?
That's what I did because, I can just got a total of 3$ if and only if I picked three 1$ coin out of 10 coins containing five 1$, two 10$ and three 5$ coins. Is my argument right?

Answer 12

Sounds good to me, but I'm not getting the same value of \(E(X)\) as you. Checking my calcuations

Answer 13

Yesd I'd say you are right yttrium.
How is that formula 5C3/10C3 set up or calculated?

Answer 14

It's a combination formula. Sorry, but I don't know it's form in factorial.
It's like n1Cr1/n2Cr2

Answer 15

okay

Answer 16

@wolf1728 , you know what I mean, right?

Answer 17

no - I don't but I figured I'd stay with the topic anyway

Answer 18

look at @SithsAndGiggles 1st comment, maybe, there you can understand what I mean.

Answer 19

After some laborious typing on a calculator, I got \(E(X)=11.125\). Would you mind showing your complete set-up?

Answer 20

Siths could you explain how this is interpreted. I know about the (10 3)

Answer 21

Uhh, maybe I can do it later, I am having work right now. Anyway, we are having almost near answeres, huh. But can I ask you one thing? Have you checked your "summation of f(x) = 1"? @SithsAndGiggles

Answer 22

Anyway, i've taken this question on our last quiz, and I just wanna make sure if I've done a right move. :3

Answer 23

@wolf1728, 2 of the cards are chosen from the 2 total of $10 coins, and 1 is chosen from the 3 total of $5 coins. (Notice that we're choosing 3 coins total). The reason they're being multiplied has to do with the \(mn\) rule.

Answer 24

Sith do you know what that formula is called?

Answer 25

@wolf1728, I'm not sure if there's a formal name for it. I think it's just an application of reasoning involving both the \(mn\) rule and conditions of independent events.

Answer 26

@Yttrium, I don't follow, what do you mean by \(f(x)=1\) ?

Answer 27

Isn't it the summation of all probabilities in this case must be equal to one? That's why you need to add all of your f(x) values, then, if it is equal to one. Your result can be considered valid. :)

Answer 28

I think you're confusing determining whether a probability density function (pdf) is valid with finding the expected value.

Given a set of probabilities \(f(x)\) for particular values of \(x\), you can say a pdf exists if
\[\sum_{\text{all }x\text{'s}}f(x)=1\]
Expected value (or the mean) is a different concept altogether, where
\[E(X)=\sum_{\text{all }x\text{'s}}x\cdot f(x)\]

Answer 29

Oh, it seems I'm missing one of the possibilities... Let's see if that's the culprit.

Answer 30

They're different concept. Yea, I agreed, but isn't it that all probabilities of all events when added will end at one.

Answer 31

Okay. :))

Answer 32

Yes, you're right, the expected value is indeed \(12\). I forgot to include the possibility of 2 $10 coins and 1 $1 coin. : )

Sorry for the confusion. And yes, all probabilities should add to 1, and they do!

Answer 33

All right yttrium !!!