Question

a box contains 4 red balls, 4 white balls, and 4 green balls. two balls are drawn out without replacement. what is the probability that both balls are the same color?

Answer 1

wrong ... my keyboard hates me

Answer 2

4/12 that is one color; and 3/11 that it is he same color

Answer 3

12/132 reduced is my final offer lol

Answer 4

what?>?

Answer 5

youre going to have to be more specific than that ...

Answer 6

Sample Space: { {r,r},{r,w},{r,g},{w,w},[w,g},{g,g} },
e.g. {r,r} = red on 1st and red on 2nd draw
P(2 balls drawn w/o repl are the same)
= P(red on 1st and red on 2nd) +
P(white on 1st and white on 2nd) +
P(white on 1st and white on 2nd), by independence (see sample space)
= 3*P(red on 1st and red on 2nd), by symmetry (*)

P(red on 1st and red on 2nd)
=P(red on 1st)*P(red on 2nd), by independence

P(red on 1st) = 4/16
P(red on 2nd) = 3/15, since there are only 3 red and 15 balls left

So, P(red on 1st and red on 2nd) = (4/16)*(3/15) and
P(2 balls drawn w/o repl are same)
= 3 * (4/16)*(3/15), using (*) above
= (1/4)*(3/5)
= 3/20

Ans. 15% chance of two same colors draw consecutively without replacement

Answer 7

for each color, you have 4 chances out of 12 to pick any of the colors

and given that you pick a certain color, you have 3 chances out of the 11 left to pick it again.

4/12 * 3/11 = 12/132, reduces to 1/11

Answer 8

ybars looks more indepth tho, so id go with that :)

Answer 9

but i count 12 balls total, whereas he typoed it up to 15 right?

Answer 10

yes, change 16 to 12 and 15 to 14.