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- 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.

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- anonymous

- schrodinger

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- kropot72

First you need to find the probabilities for 1, 2, 3 and 4 red balls in a draw of four balls, and the probabilities for 1, 2, 3 and 4 black balls in a draw of four balls.
\[\large P(1\ red)=\frac{6C1\times 9C3}{15C4}\]
\[\large P(2\ red)=\frac{6C2\times 9C2}{15C4}\]
\[\large P(3\ red)=\frac{6C3\times 9C1}{15C4}\]
\[P(4\ red)=\frac{6C4}{15C4}\]

- anonymous

what does C represents

- kropot72

For example, 9C3 means the number of combinations of 9 different things taken 3 at a time. It can also be read as '9 choose 3'.

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- kropot72

Have you studied combinations?

- anonymous

i haven't studied it yet just studying ahead of class. You getting ready for it. If you could shed some light on it. I would really appreciate it.

- anonymous

in combination so far i know is that order doesn't matter.
n!/r! (n-r)!

- kropot72

Many calculators have the combinations function. Using a suitable calculator the results for the equations that I posted are:
P(1 red) = 0.369; P(2 red) = 0.396; P(3 red) = 0.132; P(4 red) = 0.011.

- kropot72

Now you need to calculate the probabilities of 1, 2, 3 and 4 black balls in a draw of four balls.

- kropot72

\[\large P(1\ black)=\frac{4C1\times 11C3}{15C4}\]

- kropot72

Having found all the required probabilities, each value of red probability is multiplied by $10 and each value of black probability is multiplied by $15 and assigned a negative value. These products are then summed to find the expected value.

- kropot72

\[\large E(X)=\sum_{}^{}xp(x)\]

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