A basic question on Probability in quantitative aptitude I am trying some aptitude questions and this question asks for use of probability . I am not able to find the right answer.

A cupboard is filled with a large number of balls of 6 different
colours. You already. have one batl of each colour. If you are
blind-folderd, how many balls do you need to draw to be sure of having
3 colour-matched pairs of balls?
A. 3
B. 4
C. 5
D. 6

I think as there are large no of balls of each colour, so there must be $\frac{1}{6}$ probability of getting ball of 1 color and hence $12$ attempts of balls must be needed.
But I am wrong.
Answer is

 A .

Can anyone tell what mistake I am making?
 A: I'd approach it in the following way: The probability the first sampled ball is on 'new' color is obviously 1, whatever the color. Now you have one match, after the first sample. The probability to sample a ball of a different color is $\frac{5}{6}$, so the mean number of trials needed to get one of those colors is $\frac{6}{5}$. Now you have 4 'unsampled' colors left. The probability to get any of them is obviously $\frac{4}{6}$, so the mean number of samples is $\frac{6}{4}$. In total:
$$
ET = 1 + \frac{6}{5} + \frac{6}{4} = 3\frac{8}{15}
$$
I believe this is where the solution (3) comes from
A: Assuming that color-matched pairs can be of the same color:
To be sure, let's consider the worst-case of drawing balls at random.
You have $6$ balls of colors $a, b, c, d, e, f$. The worst-case goes as follows:
$(1)$ You draw $1$ ball at random and get $a$(doesn't make a difference as to which one you get on starting out).
Balls in hand so far: $aabcdef$.
Pair(s): $1(aa)$
$(2)$ You don't get one of the other balls and get an $a$ again.
Balls in hand so far: $aaabcdef$.
Pair(s): $1(aa)$
$(3)$ Either you get $a$ or one of the others (say $b$), you get another pair.
Balls in hand so far: $aaabbcdef$.
Pair(s): $2(aa,bb)$
$(4)$ You get $b$ again as the worth that could happen. Had you got $a$ in your $3^{\text{rd}}$ draw, you would have had to consider getting $a$ this time as the worst that could happen.
Balls in hand so far: $aaabbbcdef$.
Pair(s): $2(aa,bb)$
$(5)$ No matter what you draw(say $c$), you get your last pair.
Balls in hand so far: $aaabbbccdef$.
Pair(s): $3(aa,bb,cc)$
Hence, $5$ attempts to be sure.
PS: I don't know how the answer is option A considering my assumption and that one has to be sure. If it was an "average-attempts to be sure(kinda)" question, I would have approached it by Probability.
