anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
b) asked for pmf P_x (8/14)^2 x=-2 (4/14)*(8/14) x=1 (4/14)^2 x=2
anonymous
  • anonymous
@zarkon
anonymous
  • anonymous
@myininaya

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More answers

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

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