Question

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 $6, and for each black ball drawn, you lose $9. Let X represent your net winnings.

Answer 1

what do u need to find ?
expected profit ?

Answer 2

That's correct. Sorry, I should have put that in there.

Answer 3

Can you find the probability of selecting a red ball from the urn ?

Answer 4

I also need some more info.
how many times in total, you will be drawing the ball ?

Answer 5

ok, 4 times

Answer 6

Only once with no replacement I believe.

Answer 7

Find total number of balls in the urn .

P(red ball) = total number of red balls / total number of balls in the urn = ...

Answer 8

Oh, you meant how many times a single ball is drawn. Then yes, you're correct. 4 times.

Answer 9

P(6)=6/15=.4

Answer 10

good, so you know how to calculate probabilities.

ok, in first draw, there are 15 balls,
you can pick a red OR a black ball.

so your expected profit in 1st draw will be

P(red)*6 - P(black)*9

The second term is subtracted as you lose $9 if you draw a black ball

does that make sense ?

Answer 11

P(black)=4/15=.3 correct?

Answer 12

yes
bu note that what i stated is true only for one(first) draw

Answer 13

Get the probability of drawing 4 red balls in 4 draws of balls from the urn without replacement

Answer 14

There are 15 possible combinations of red and black, as follows:
R B
4 0
3 0
2 0
1 0
0 0
3 1
2 1
1 1
0 1
2 2
1 2
0 2
1 3
0 3
0 4
For starters, the probability of each needs to be found.

Answer 15

Wait, isn't this where I would do something like
4/15 * 3/14 * 2/13 * 1/12 ?

Answer 16

As you know 0.4 is just the probability of drawing one red ball from the urn in one draw.

Here's what you need to do,

Take a row from the above table krotop posted, say 4,0
'Get the probability of drawing 4 red balls in 4 draws of balls from the urn without replacement'
which is what i asked.

and your reply
'4/15 * 3/14 * 2/13 * 1/12'
for this is correct :)

Repeat this for all the rows
mentioned in that table

Answer 17

ok

Answer 18

This is going to take a minute. LOL. I know there's a quicker way to do this using the nCr function on my calculator, but I've been doing this stuff all day and I'm starting to confuse myself. lol

Answer 19

Once you get the probabilities, multiply it with your profit!

like for the first row, your profit will be 4*6 =$24
multiply 24 with the probability of 1st row


of you take the row with 1,3
1 red and 3 balck = +6 - 3*9 = -21
multiply -21 with the probability you calculated for 1,3

repeat for all rows, and add all the profits

Answer 20

do you have some idea on what you need to do /

Answer 21

Oh ok. Gotcha. Just doing the math now.

Answer 22

Just to make sure I'm doing this right before I get too far, the answer for the first row would be 83865.6?

Answer 23

For the first row, the probability of drawing 4 red is 0.01099. Then you multiply that value of probability by $24.

Answer 24

Ok, I was doing it wrong. Didn't get 0.01099

Answer 25

The probability of drawing 4 red is given by:
\[\large P(4\ red)=\frac{6C4}{15C4}\]

Answer 26

Ok, now I got 0.01099.

Answer 27

The same would be for the second row, but instead I would use P(3red)=6C3/15C4 correct?

Answer 28

The probability of drawing 3 red, 1 white and 0 black is given by:
\[\large P(3\ red,\ 1white,\ 0\ black)=\frac{6C3 \times 5C1}{15C4}\]

Answer 29

Yep, gotcha. Thanks.

Answer 30

yw

Answer 31

the very last row is giving me an answer with \[\epsilon ^{-4}\]

Answer 32

.0282 is what I got as my final answer. Hoping that was right.

Answer 33

The probability of 4 black and 0 red is given by:
\[\large P(4\ black)=\frac{1}{15C4}=0.000733\]