anonymous
  • anonymous
An urn contains six red balls, five white balls, and four black balls. Four balls are drawn from the urn at random without replacement. For each red ball drawn, you win $10, and for each black ball drawn, you lose $15. Let X represent your net winnings. Compute E(X), your expected net winnings.
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Hya
anonymous
  • anonymous
hello
anonymous
  • anonymous
I'm not good at this :P

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anonymous
  • anonymous
I'll give you a medal :D
anonymous
  • anonymous
I need help not medal lol
anonymous
  • anonymous
:P
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
i cannot think of a snappy way to do this, this is going to take like forever
misty1212
  • misty1212
first you have to think of all the possible outcomes when you pick 4 balls
misty1212
  • misty1212
then you have to compute the amount of money you win or lose for each case then you have to find the probability of each possible outcome then you have to multiply and add !
anonymous
  • anonymous
could you lead me through the problem step by step please.
misty1212
  • misty1212
if i had like two hours i could lets at least begin
anonymous
  • anonymous
sure thing thanks
misty1212
  • misty1212
one possibility is you get all red and therefore win $40 the probability you get all red is \[\frac{\binom{6}{4}}{\binom{4}{15}}\]
misty1212
  • misty1212
ok that is not quite right, the probabilty is \[\frac{\binom{6}{4}}{\binom{15}{4}}\]
misty1212
  • misty1212
that is \(\frac{1}{91}\) http://www.wolframalpha.com/input/?i=%286+choose+4%29%2F%2815+choose+4%29
misty1212
  • misty1212
then you could have 3 red, one white, win $30 that probability is \[\frac{\binom{6}{3}\times \binom{5}{1}}{\binom{15}{4}}\]
misty1212
  • misty1212
this is really just a start there are lots of other possibilities i can't think of a quick way to do this though
anonymous
  • anonymous
so if i get this right this is how it goes. P(1 red)=6C1×9C315C4 P(2 red)=6C2×9C215C4 P(3 red)=6C3×9C115C4 P(4 red)=6C415C4 P(1 red) = 0.369; P(2 red) = 0.396; P(3 red) = 0.132; P(4 red) = 0.011.
anonymous
  • anonymous
the problem gets tough here on out i don't know what to do from here
anonymous
  • anonymous
P(1 red)=6C1×9C3/15C4 P(2 red)=6C2×9C2/15C4 P(3 red)=6C3×9C1/15C4 P(4 red)=6C4/15C4 correct way
anonymous
  • anonymous
@kropot need your help from this part
anonymous
  • anonymous
i didn't understand anything from here on out \
kropot72
  • kropot72
P(1 black) = 0.484; P(2 black) = 0.242; P(3 black) = 0.32; P(4 black) = 0.001 E(X) = 10(0.369 + 0.396 + 0.132 + 0.011) - 15(0.484 + 0.242 + 0.032 + 0.001)
kropot72
  • kropot72
E(X) = a loss of $2.41.
anonymous
  • anonymous
@kropot72 but we need to find the net winnings not loss
kropot72
  • kropot72
Expected net winnings = -$2.41.
kropot72
  • kropot72
"Net winnings" means (expected gains) - (expected losses). In this case the expected losses outweigh the expected gains.
anonymous
  • anonymous
give me medal?

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