Question

There are 3 red sticks, 4 yellow sticks, and 2 blue sticks in a bag. Sticks are picked one by one. If the stick is red then it goes back in the bag, but if it is not red, then it is kept out of the bag. You stop picking after 3 such picks.

Y is the random variables that counts the # yellow sticks - #red sticks. Find the expected value of Y.

Answer 1

@jim_thompson5910

Answer 2

first we need all possible values of Y, then we need the probability of each value. i believe this will take a couple minutes, although there might be a short cut

Answer 3

Isn't the number of yellow sticks just 4 and red sticks 3?

Answer 4

it seems like Y can take on the values
\[\{-3,-2,-1,0,1,2\}\]

Answer 5

How did you get those?

Answer 6

Y is the number of yellow sticks selected, minus the number of red sticks selected, out of the total of 3 sticks selected

Answer 7

so for example, if you select no yellow stick and all red sticks, then \(Y=0-3=-3\)

Answer 8

But, in order to get 2 it would be 5-3, but there aren't 5 yellow sticks

Answer 9

now for each of these you need the associated probability
and for that you need to figure out exactly what would give you each number

to answer your last question, if you select 2 yellow and no red, then \(Y=2-0=2\)
or if you select 3 yellow and 1 red then \(Y=3-1=2\)

Answer 10

i interpret the statement
"Y is the random variables that counts the # yellow sticks - #red sticks. " to mean out of the three selected

Answer 11

our next job is to figure out each of the associated probabilities. i.e. \(P(Y=-3),P(Y=-2), P(Y=-1)...\)

Answer 12

Alright, so if I picked 3 yellow sticks and 1 red, would that be 4*3*2*2?

Answer 13

lets start at the bottom
\(P(Y=-3)\) means all 3 are red
that probability is \(\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{27}\)

Answer 14

is that ok?

Answer 15

Yep, that looks right

Answer 16

how about \(PY=-2\) that means either no yellow and two red or one yellow and 3 red

Answer 17

no i don't think so
this problem is a real pain because of the different conditions
replacement and non replacement
lets compute the probability of no yellow and two red.

Answer 18

no yellow and two red would be 1/3 * 1/3 right?

Answer 19

i think no yellow and two red is
\[3\times \frac{1}{3}\times \frac{1}{3}\times \frac{2}{9}\]

Answer 20

r r b
r b r
b r r

Answer 21

I get the (1/3)*(1/3)*(2/9) for blue, but where did the 3 come from?

Answer 22

three ways to get it

Answer 23

That makes sense!

Answer 24

ok i think that is the only way for \(Y=-2\) because we can't have one yellow and 3 red if we only pick 3

Answer 25

now on to \(Y=-1\) and this is going to be a pain i think.
1 yellow 2 red or
no yellow 1 red

Answer 26

no yellow one red is \(3\times \frac{2}{9}\times \frac{1}{8}\times \frac{1}{3}\) i think but i could be wrong

Answer 27

b b r
b r b
r b b

Answer 28

no it is wrong damn

Answer 29

b b r: \(\frac{2}{9}\times \frac{1}{8}\times \frac{3}{7}\)

Answer 30

b r b: \(\frac{2}{9}\times \frac{3}{8}\times \frac{1}{8}\)

Answer 31

this is really a pain
ok finally
r b b: \(\frac{3}{9}\times \frac{2}{9}\times \frac{1}{8}\)

Answer 32

reason being if you pick a red first you replace it, so there are still 9 sticks in the bag

Answer 33

So, I would add those up to get the probability of Y=-1?

Answer 34

well believe it or not, no, because that was only for two blue and one red
you also have to do 1 yellow 2 red
then add them
i have to run, but i hope it is clear what you have to do
btw ignore \(Y=0\) since this has nothing to do with the expected value, as when you multiply by 0 you get 0

Answer 35

I drew out a tree diagram and for 1Y 2R I got:
(1/3)*(4/9)*(3/8) ...RYR
(1/3)*(1/3)*(4/9) ...RRY
(4/9)*(3/8)*(3/8)... YRR
So I would multiply and add these right?

Answer 36

RBB=(1/108)
BRB=(1/96)
BBR=(1/84)
RYR=(1/18)
RRY=(4/81)
YRR=(1/16)
Would I just add all these up?

Answer 37

If I just add up the values, then...
P(Y=-3)=(1/27)
P(Y=-2)=(2/27)
P(Y=-1)=(.199)
P(Y=0)=0 you said?
P(Y=1)=(238/1008)
P(Y=2)=(1/7)

What would I do now?

Answer 38

assuming those probabiities are correct, and i didn't check them, you multiply and add

Answer 39

\[-3\times P(Y=-3)-2P(Y=-2)-1P(Y=-1)\]
\[+P(Y=1)+2P(Y=2)+3P(Y=3)\]

Answer 40

So, I went back to make sure everything was right and for Y=-2 I got
(1/3*)(1/3)*(2/9)= RRB
(1/3)*(2/9)*(3/8)=RBR
(2/9)*(3/8)*(3/8)=BRR

But that's not what you got. Could you just check just that part really quickly since I want to make sure I didn't mess everything up

Answer 41

that looks correct

Answer 42

two reds, one blue

Answer 43

Thanks! Also, do you know how to find the density function? I'm not sure what that is

Answer 44

this is a pain this problem
but what you wrote looks good

Answer 45

this is exactly what you are doing. you are finding the probability density function for this random variable \(Y\)

Answer 46

I thought I was finding the expected value?

Answer 47

this random variable takes on the values \(\{-3,-2,-1,1,2,3\}\) and finding the probability for each of these values is exactly the probability density function
it is not the expected value, but you need this first

Answer 48

the expected value is what you get when you multiply and add

Answer 49

Alright, so {−3,−2,−1,1,2,3} is the density function?

Answer 50

i wrote it a few posts above

Answer 51

no the density function is not the possible values of \(Y\)
it is this \(f(x)=P(Y=x)\) it only has 6 elements in the domain of the function

Answer 52

so for example \(f(-2)=P(Y=-2)\) which is what you computed above

Answer 53

So, it is the probabilities of each of the Y=...?

Answer 54

yes

Answer 55

Thank you so much!

Answer 56

\(f(-1)=P(Y=-1)\) etc

Answer 57

yw
you still have to multiply and add right?

Answer 58

Wait, is it the probability of Y * the number? Like -1*P(Y=-1) or
3*P(Y=3)?

Answer 59

\[-3\times P(Y=-3)-2\times P(Y=-2)-P(Y=-1)+P(Y=1)+2P(y=2)+3P(y=3)\]

Answer 60

multiply and add over all possible values of \(Y\)

Answer 61

do you have these probabilities?

Answer 62

Yep, I finished the problem awhile ago. I just needed to know what the density function is for a different problem.

Answer 63

discrete or continuous?

Answer 64

It just find the density function. I'm a bit confused though. You said it was just P(Y=..), but then you posted ...*P(Y=...). Is the density function the probabilities of Y * the number in from like -1*P(Y=-1) or 3*P(Y=3)?

Answer 65

*says *front

Answer 66

ok it is the expected value that you multiply and add

Answer 67

So, the density function is just the table of Y=... without multiplying Y=... by the number in front, right?

Answer 68

not the density function. but if you rewrite \(P(Y=x)\) as \(f(x)\) then the expected value is \[\sum_{x}xf(x)\]

Answer 69

Thank you!

Answer 70

yes, the density function is the the values \(f(x)=P(Y=x)\)
just the values
the expected value is what you get when you multiply and add
yw