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.

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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
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Hya
hello
I'm not good at this :P

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Other answers:

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

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