Question

I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different balls at random. What is the expected value of the sum of the numbers on the marbles?

Answer 1

it is the some of what you get times the probability you get it

Answer 2

But since it asks for the sum, how would I do that part?

Answer 3

you can add i believe
lets try it

Answer 4

the expected value of one pick is
\[\frac{1+2+3+4+5}{5}=3\]

Answer 5

so my guess is the expected value of two picks is 6

Answer 6

How would you find that though?

Answer 7

we could also work it out the long way, probably get 6 that way too

Answer 8

put \(X\) as the sum of the picks
then find the possible values of \(X\)
could be \(X=2,X=3,X=4,X=5,X=6,X=7,X=8,X=9,X=10,\)

Answer 9

find the probability of each, then multiply and add them up

Answer 10

oh that is wrong sorry, you can't get some of these
lets be more careful

Answer 11

you can't get 2 and you can't get 10

Answer 12

you can get a total of 3, that probability is \(\frac{1}{10}\)

Answer 13

you can get 4, that probability is also \(\frac{1}{10}\)

Answer 14

you can get 5 and that probability is \(\frac{2}{10}\)

Answer 15

is it clear where i am getting these numbers from?

Answer 16

sorry, just a sec

Answer 17

k

Answer 18

I'll just message you when I'm back

Answer 19

yeah

Answer 20

how far did you get ?

Answer 21

"it is the some of what you get times the probability you get it "

lol "the some" :D

Answer 22

Let me find my notebooks, just a sec

Answer 23

the some
did i write that?

Answer 24

oh yeah! lol

Answer 25

I gave you a medal cos I'm cracking up haha.

Answer 26

it is "some of what you get" sometimes

Answer 27

Annd I've managed to lose most of the work i've done.... smh

Answer 28

Also, I'm wondering, in the probabilities, is something like 1,3 different from 3,1?

Answer 29

ok if you want the expected value the first step is to write down a random variable, because you have to take the expected value of something

Answer 30

in this case you want the expected value of the sum
so put \(x=sum\)

Answer 31

okay,

Answer 32

the put what the possible values of \(x\) can be
in this case \(x\) could be
\(\{3,4,5,6,7,8,9\}\)

Answer 33

1,3 and 3,1 means there's 2 diff. ways of getting 4, which increases the probability of getting 4. But you shouldn't have to worry about that, cos that applies to all combos (eg 1,5 and 5,1, 3,2 and 2,3 etc(

Answer 34

the you have to figure out what the associated probabilities are

Answer 35

5 choose 2 is 10, so we can use that as the denominator for all of these
that way we don't have to worry about 14 and41 for example

Answer 36

Oh, okay, that makes more sense now. So then the numerator would just be the number of distinct pairs that add to each sum.

Answer 37

\(P(x=3)=\frac{1}{10}\) there is only 1 way to get a 3, namely a 1 and a 2

Answer 38

yes, if you use 10 as your denominator, you only need to count how many ways, not order

Answer 39

1 way to get a 4: 1 3
2 ways to get 5: 1,4 or 2, 3
2 ways to get 6: 1, 5 or 2, 4

Answer 40

etc

Answer 41

i think it is 1, 1, 2, 2, 2, 1, 1 respectively for 3, 4, 5, 6, 7, 8, 9
notice that they add up to 10 which is good

Answer 42

For 6, I got 3,3 1,5 and 2,4

Answer 43

final job is to multiply and add (find the SUM)

Answer 44

think carefully here

Answer 45

Ohh, I forgot, no replacement.

Answer 46

right

Answer 47

I got... 6

Answer 48

Is it 6? I might have messed up in my calculations.

Answer 49

probably is 6 because we got 6 without thinking at the beginning

Answer 50

yeah it is six
the quick way to do it is put \(x\) as the number chosen first, the \(E(x)=3\) and \(y\) as the second number chose so \((Ey=3\) then the expected value of the sum is \(E(x=y)=Ex+Ey=6\)

Answer 51

How do you know the expected value of the second number chosen is 3?

Answer 52

Probably because it's equally probable that you get a 5, leaving 1,2,3,4, or get a 1, leaving 2,3,4,5... all those possibilities are equal and likely lead to the expectation of the second ball being 3.

Answer 53

because it is symmetric

Answer 54

first second
second first
you cannot tell them apart

Answer 55

ie it's equally as likely to have 1,3,4,5 left as 1,2,3,5 etc.

Answer 56

(after picking the first ball)

Answer 57

plus, there is no real "first second" anyway
the problem says "pick two" you can put two hands in and pick if the bag is big enough

Answer 58

like the difference between tossing two coins once or tossing one coin twice

Answer 59

Ah, I see. Also, 6 is the final answer, correct? Or am I missing a step.

Answer 60

6 is correct.

in fact the expectation will be 3 times however many balls you pick out. If you pick out all 5, the sum is 15... 5*3.

Answer 61

I'ma just give the medal to agent0smith... considering 100 is pretty much as high as you can go.

Answer 62

But thanks to both of you guys.

Answer 63

just out of curiosity
how can you pick 15 balls out of a bag that contains only 5 balls?

Answer 64

i tried that once, taking $15 dollar bills out of a wallet that only contained 5,
didn't work though

Answer 65

ooh i see, pick 5 out of 5
must be seeing things....

Answer 66

Hmm, it appears that the expectation is the same regardless of whether it's with replacement or not... the expectation on every ball is 3, it's just that with replacement you can pick over 5 balls.

Answer 67

lol "the some" if you pick 5 balls is 15, @satellite73 ;)