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

I assume you mean on consecutive picks...
a. 1/9 is the closest
you have a 4/10 probability multiplied by a 3/9 probability.
You wouldn't add the possibilities as done a few answers above because that does not make sense you may average them instead to receive the accurate answer.

Answer 2

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 3

When you grab the first marble, the chances of it being a red one are 1/10. The chances of you picking another red marble are then 3/9. The overall probability of your selecting 2 red marbles in a row is 3/90 (the product)

Do the same for the blue and yellow marbles and add all of them up. That's the total probability of getting 2 with the same color.

So 3/90 + 2/90 + 1/90 = 6/90 = 1/15

The correct answer isn't given in your list. (And I have *no* idea what answer d. could even mean)


Doug

Answer 4

2/9

Answer 5

didn't you already ask this question with red balls?
like I said with that question,
10 balls and lets say 2 yellow.
with the first pick, 2/10 chance because there are 2 yellow in 10 balls.
second pick
9 balls left, 1 yellow left, so 1/9
so 2/10 x 1/9

this question is stupid because you can't remove two white balls.

Answer 6

The answer is b, 2/9.

Answer 7

(4C2+3C2+2C2)/10C2

=( 6 + 3 + 1 )/ 45

=10/45=2/9

Answer 8

20%

Answer 9

c.
Whats the answer, not sure.

Answer 10

B