Libniz
2 balls are chosen randomly from an urn containing 8 white, 4 black, 2 orange ball. Suppose you win 2 dollars for each black ball and lose 1 dollar for each white ball selected. Let X be the net winning , Find
a) S_x
my answer is
for a)
(-2,1,2)
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Libniz
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b) asked for pmf
P_x (8/14)^2 x=-2
(4/14)*(8/14) x=1
(4/14)^2 x=2
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@zarkon
Libniz
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@myininaya
MathLegend
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What type of math is this? Algebra 1?
MathLegend
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Honestly, I have no clue.
OpenstudyUser
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lol
Libniz
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probability
Zarkon
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sampling with or without replacement?
Libniz
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didn't say anything about replacement so I think without
Zarkon
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if it is without...then your pmf is not correct
Libniz
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what is it mean by "replacement"?
Zarkon
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if it is replacement then you pick a ball...look at it ..put it back then randomly sample again with all 14 balls
Libniz
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I think ,from the wording; we are pulling two balls out at the same time
Zarkon
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that is what I am thinking
Zarkon
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so you need to redo your pmf
Zarkon
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if you get black balls then you get $2 each correct
Libniz
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yes
Zarkon
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so one possibility is that you win $4
Libniz
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white ball, I lost $1
Zarkon
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BB=$4
BO=$2
OB=$2
BW=$1
WB=$1
...
Zarkon
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so X can take the values 4,2,1,0,-1,-2
Zarkon
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agree?
Libniz
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oh, I see, I omitted orange which I shouldn't have
Libniz
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so how would I write PMF?
Zarkon
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what i s the probability of 2 blacks? ie $4
Libniz
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(8/14)*(7/13)
Zarkon
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ok
what about winning $2
Libniz
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(4/14)(2/13)
or
(2/14)(4/13)
?
Zarkon
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winning $2 is BO or OB
Zarkon
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there are 8 blacks...i see no 8's in your solution
Zarkon
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lol.....there are 8 W
Zarkon
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4 black...so prob of BB is
\[\frac{4}{14}\cdot\frac{3}{13}\]
Zarkon
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not what you had above
Zarkon
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so \[P(X=4)=\frac{4}{14}\cdot\frac{3}{13}\]
Libniz
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right , so for $2
(4/14)(2/13)
or
(2/14)(4/13)
(4/14)(2/13)+(2/14)(4/13)=
Zarkon
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yes
Zarkon
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the sum of the two
Libniz
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I understand now; I really appreciate your help
Zarkon
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no problem