# E(X), bag with 6 red and 5 black balls, X = number of red balls

#### Hans Rudel

##### New Member
A bag contains 5 black and 6 red balls. Two counters are drawn, one at a time, and not replaced. Let X be "the number of red counters drawn". Find E(X).

ive drawn up a tree diagram which im assuming is correct

EDIT: Below is the corrected tree diagram, as eagle eye Dason had spotted a mistake in the original.

from this i then calcd that (Corrected)

P(X=0) = 2/11
P(X=1) = 6/11
P(X=2) = 3/11

and E(X) = (0 x (2/11)) + (1 x (6/11)) + (2 x (3/11)) = 12/11 (Corrected due to mistake above)

(Should be correct now) Id appreciate it if someone could point out where i have made a mistake.

Thanks

Last edited:

#### Dason

Re: E(X), bag with 5 red and 6 blue balls, X = number of red balls

Your tree diagram isn't correct. Look at the first branch and ask yourself which color should have a higher probability.

#### Hans Rudel

##### New Member
Re: E(X), bag with 5 red and 6 blue balls, X = number of red balls

hmm D'oh. Hopefully that sorts it out.

#### Dason

Re: E(X), bag with 5 red and 6 blue balls, X = number of red balls

Hopefully that's the problem? I wasn't positive because your question title and the problem as stated in the question contradict which of the colors have 5 and which has 6.

#### Hans Rudel

##### New Member
Re: E(X), bag with 5 red and 6 blue balls, X = number of red balls

I get the correct answer now so, you were right. Well spotted + thanks again for your help.

Regarding the question, hmm i guess i need to take a break. Is it possible to edit the thread title? If not i will just edit the first post so that it clearly states that i had made a mistake.