# Math Help - stuck on probability problem

1. ## stuck on probability problem

I'm stuck on this question. It is an expected value question involving marbles:

"We have 20 marbles in a bag: 17 red and 3 blue. We randomly pull three marbles without replacement. If we pull three red marbles, we pay $10. In any other case we receive$10 for every blue marble pulled. (1 blue gets $10, 2 blue gets$20....) What is the expected value of this game?

I'm thrown off by the third marble.

2. Hey

Originally Posted by jlefholtz
I'm stuck on this question. It is an expected value question involving marbles:

"We have 20 marbles in a bag: 17 red and 3 blue. We randomly pull three marbles without replacement. If we pull three red marbles, we pay $10. In any other case we receive$10 for every blue marble pulled. (1 blue gets $10, 2 blue gets$20....) What is the expected value of this game?

I'm thrown off by the third marble.
You need to find the probabilities first

p("three red") = $\frac{17}{20}*\frac{16}{19}*\frac{15}{18}$
p("one blue") = prob(blue,red,red)+p(red,blue,red)+p(red,red,blue) $= \frac{3}{20}*\frac{17}{19}*\frac{16}{18} + \frac{17}{20}*\frac{3}{19}*\frac{16}{18}+\frac{17} {20}*\frac{16}{19}*\frac{3}{18}$

(I hope in this formula is no typo in it)

$p("two blue") = p(red,blue,blue)+p(blue,red,blue)+p(blue,blue,red)$

p("three blue") = p(blue,blue,blue)

I think you can solve it yourself

and then E(x) = -10*p(three red) + 10*p(one blue) + 20*p(two blue)+30*p(three blue)

Rapha

3. ## the answer i got.

This is what I did, I'm not sure if I did it correctly though.

P (3 red) + P (3 blue) + P(2 red, 1 blue) + P (2 blue, 1 red)=

(17/20 x 16/19 x 15/18)(-$10) + (3/20 x 2/19 x 1/18)($30) + (17/20 x 16/19 x 3/18)($10) + (17/20 x 3/19 x 2/18)($20)= -$4.44 4. thanks rapha! by the looks of the reply you gave me, it looks like i did it right. thank you for your reply. 5. Originally Posted by jlefholtz This is what I did, I'm not sure if I did it correctly though. P (3 red) + P (3 blue) + P(2 red, 1 blue) + P (2 blue, 1 red)= (17/20 x 16/19 x 15/18)(-$10) + (3/20 x 2/19 x 1/18)($30) + (17/20 x 16/19 x 3/18)($10) + (17/20 x 3/19 x 2/18)($20)= -$4.44
Im not sure about it, because i got

P (3 red)*(-10) + P (3 blue)*(30) + P(2 red, 1 blue)*(10) + P (2 blue, 1 red)*(20)=

17/20*16/19*15/18*(-10) + 3/20*2/19*1/18*(30) + 3*(17/20*16/19*3/18)*(10) + 3*(3/20*2/19*17/18)*20

= - 1.47

You forgot 3* P(2 red, 1 blue) ... 3* P(2blue, 1 red)

because there are three different possibilities, (blue, red, red), (red,blue,red), (red,red,blue). Thus 3*p(blue,red,red) = p(1blue, 2 reds)

6. ## oops

oh, wow, thanks for catching that. i didn't check my work. thanks again.