A woman selects balls from a bowl at random … 
A bowl contains 10 red balls, 10 blue balls and 10 black balls. A woman selects balls at random without looking at them:
a) How many balls must she select to be sure of having at least three balls of the same color?
b) How many balls must she select to be sure of having at least three blue balls?

My solutions:
a) $N=30$, $k=2$, $[N/2]\ge 3$ so $N=2\cdot(3-1)+1= 5$
b) 23, because if she select 10 red balls and 10 black balls, than she must select those 3 blue balls to be sure that they are at least three of them blue..

Second question:

A bowl contains 15 red balls and 15 black balls. A woman selects balls at random without looking at them:
a) How many balls must she select to be sure of having at least three balls of the same color?
b) How many balls must she select to be sure of having at least three black balls ?

My solutions:
a) $N=30$, $k=2$, $[N/2]\ge 3$ so $N=2\cdot (3-1)+1= 5$
b) $18$, because if she select 15 red balls, than she must select those 3 black balls to be sure that they are at least three of them black..
Please correct me if I'm wrong!
 A: The solution to a) is 7 by the Pigeon hole principle. After selecting 6 balls, the only way to not have 3 of the same colour is to have selected 2 of each colour, at which point selecting a 7th ball will give the 3rd ball of a certain colour.
The solution to b) is 23 by the pigeon hole principle. You found that correctly.
Similarly for the second question:
The solution to a) is 5 as we could select 2 of each, but then the 5th ball will give us 3 of the same colour.
The solution to b) is 18 as we could select all 15 red balls, but then we must select 3 black balls in the next 3 balls.
A: I don't agree with your first number: the only way to avoid 3 equals in 6 moves is 2/2/2 and after that we always are done, so we need 7.
With 3 blues, $23$ seems correct too; worst case 10/10/0 but then there are only blues left. So 22 does not suffice, 23 does.
As to the 15 15 urn: 5 is then correct for 3 equals and 18 for 3 blacks, in similar reasoning.
A: a)Consider the worst case in which she will choose 2 balls of each colour first ,and then the third ball.So there will be 2×3+1 balls needed.so answer is7
b)again the worst case in which she will choose 10 red balls and 10 black balls and then the 3 blue balls .so answer is 23
Again in the second question
a)She will choose 2×2+1=5
b)worst case in which she chooses 15 red balls and then the 3 black balls =18 balls
