10 balls( 4 red,3 blue,4 yellow,1 wthite). remove 2 balls. probability for remove 2 balls with same colour?

Question

a.1/9 b.2/9 c.2/3 d.10

Answer 1

Choice d. is weird...
P(2 same color) =
P(2 reds) + P(2 blues) + P(2 yellow) + P(2 white) =
(4/12 * 3/11) + (3/12 * 2/11) + (4/12 * 3/11) + (1/12 * 0/11)
= 2/22 + 1/22 + 2/22 + 0 = 5/22
HUH?
Wait a minute! You says there are 10 balls, but 4+3+4+1=12!

Answer 2

Your problem appears logically inconsistent. You state there are a total of 10 balls but the sum of 4 red, 3 blue, 4 yellow and 1 wthite balls is 12 balls. If you begin with 10 balls then you need to restate your parsing of ball colors.

Answer 3

5 to 12

Answer 4

PROBABILITY OF 2 RED BALLS IS 4/10*3/9=12/90=2/15

SOLVE FOR OTHER COLORS AND ADD THE RESULTS..

Answer 5

assuming initially 12 balls, and after the removal of two,we have 10 balls
(p for 2 same color balls)=(5/22) =0.227 ~=0.23

which is closer to answer B (2/9 = 0.22)

Answer 6

remove 2 red

Answer 7

p(red1) = 4/12 p(red2) = 3/11
p(blu1) = 3/12 p(blu2) = 2/11
p(yel1) = 4/12 p(yel2) =3/11
p(whi1) = 1/10 p(whi2) = 0

p two balls = p(red1)*p(red2) + p(blu1)*p(blu2) + p(yel1)*p(yel2) + p(whi1)*p(whi2)

p two balls = 12/132 + 6/132 + 12/132 + 0 = 30/132 = .227

about 2/9

Answer 8

1p9 for the first ball + (1p8 ) for the 2nd one,, if you are removing 1 after another

Answer 9

Check your addition first.....

4 red
3 blue
4 yellow
1 white
____________
12 balls

Ya have to have the premise correct before you start solving for probabilities....

Answer 10

Ummm... you've given us 12 colors.