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

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

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