10 balls( 4 red,3 blue,2 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

First check 4+3+2+1=10 OK!

Choice d. is still weird...
P(2 same color) =
P(2 reds) + P(2 blues) + P(2 yellow) + P(2 white) =
(4/10 * 3/9) + (3/10 * 2/9) + (4/10 * 3/9) + (1/10 * 0/9) =
12/90 + 6/90 + 12/90 + 0 = 20/90 =2/9

Answer 2

Considering each ball of the same colour as a different ball since they're indistinguishable, there are 9+8+7+6+5+4+3+2+1 = 45 different combinations. Since there are 4 reds, there are 3+2+1 = 6 different combinations when there are 2 reds. For blue, it's 3, and for yellow, it's 1. That's a total of 10/45, which is 2/9.

Answer 3

Red: (4/10) * (3/9) = 2/15
Blue: (3/10) * (2/9) = 5/90
Yellow: (2/10)* (1/9) = 2/90
White: (1/10) * (0/9) = 0

Add these fractions together and we get alomost 21% chance of removing balls of the same color. The closest answer to that is 2/9, so that is my guess.

Answer 4

2/9

Answer 5

10 balls( 4 red,3 blue,2 yellow,1 wthite). remove 2 balls. probability for remove 2 balls with same colour?
Solution:
{(4C2)+(3C2)+ (2C2)}/ [10 C2}
= {(6)+(3)+ (1}/ [45}
= 10/45
= 2/9

Answer 6

1/9........

differs...for each colour.

Answer 7

2/9 ... BUT U WON'T LEARN ... HOW TO DO IT...AND THAT WOULD REALLY SUCK...