An urn contains six red balls, five white balls, and four black balls. Three balls are drawn from the urn at random without replacement. For each red ball drawn, you win $4, and for each black ball drawn, you lose $6. Let X represent your net winnings.

Compute E(X), your e

Question

An urn contains six red balls, five white balls, and four black balls. Three balls are drawn from the urn at random without replacement. For each red ball drawn, you win $4, and for each black ball drawn, you lose $6. Let X represent your net winnings.

Compute E(X), your e

jtvatsim · Answer

E(X) is your expected value, correct?

perl · Answer

if you draw all red balls, you get 4*3 winning

perl · Answer

|dw:1417569225985:dw|

perl · Answer

so lets look at possibilities

RRR (win 12 dollars)

RRB ( win 8 lose 6 , so 2 ) 

RRW

anonymous · Answer

perl thank you again but what is the final answer

perl · Answer

there are many possibilities to consider

anonymous · Answer

yes i did it but there is a loss of 46 dollar

anonymous · Answer

i tried but the answer is wrong.

perl · Answer

can you make me a chart of all possibilities

RRR
RRB
RBR
...
WWW

perl · Answer

all possible ways of drawing three balls, given that you can pick Red, black or white

anonymous · Answer

it is a long one i made it but the answer is wrong..

perl · Answer

that way I am sure we have covered all the cases

perl · Answer

can you paste it here

anonymous · Answer

ok

anonymous · Answer

RRR
RRW
RRB
RWR
RWW
RWB
RBR
RBW
RBB
WRR
WRW
WRB
WWR
WWW
WWB
WBR
WBW
WBB
BRR
BRW
BRB
BWR
BWW
BWB
BBR
BBW
BBB

perl · Answer

I will be back, brb

anonymous · Answer

THANK YOU BROTHER

kropot72 · Answer

A draw that results in either winning or losing money must have at least one red or at least one black ball. Therefore the combinations that meet this requirement are: 
1. 3 red (win $12) 
2. 2 red and 1 white (win $8). 
3. 1 red and 2 white (win $4) 
4. 3 black (lose $18) 
5. 2 black and 1 white (lose $12). 
6. 1 black and 2 white (lose $6). 
7. 2 red and 1 black (win $2). 
8. 1 red and 2 black (lose $8). 
9. 1 red and 1 black and 1 white (lose $2). 
If we let the probabilities of the nine combinations be P(1), P(2), ............... , P(8), P(9) the expected value of X is given by: 
E(X) = {P(1) * 12} + {P(2) * 8} + {P(3) * 4} - {P(4) * 18} - {P(5) * 12} - {P(6) * 6} + {P(7) * 2} - {P(8) * 8} - {P(9) * 2} .................(1) 
Now you need to find the value of probability for each of the 9 combinations and plug them into equation (1).