Submitted by Somesh Patil

To make sure Bella gets a red pen ( at least one), we must suppose the worst case .

The worst case is, Bella is not so lucky to pull out a red pen immediately, she pulls out all the blue pens and black pens. However, the next pen that she will pull out will be a red pen .(Because there are no pens in other colours in the drawer)

As stated in the problem, the drawer has $18$ blue pens and $12$ black pens. There are $18+12=30$ blue and black pens altogether. Therefore, the minimum number of pens she must pull out $=30+1=\boxed{31}$ .

In the worst case scenario, Bella would have to pull out all of the blue and black pens to get a red one.

Because there are $30$ non-red pens, Bella will have to pull out a minimum of $30 + 1 = 31$ pens out of drawer to be one-hundred percent sure she has taken out a red pen.

Worst situation, she has pulled out only black and blue pens. Finally, she pulled out pens of other color, in the box there are only red pens. Answer is: 12+18+1=31... She must pull out 31 pens. :)

We need to take here extreme cases that is we consider whatever we take, we first take out the blue and the black pens out(if possible). $18$ pens $+12$ pens $=30$ pens, what I mean is that if our answer is $30$ ,then we can claim that those $30$ belong to only blue and black colours,so our answer is $31$ ,why?Since there are no more pens of colour other than red.Hence, our answer is $\boxed{31}$ .

1. To get the possibility that red pens are pulled out, we must eliminate the possibility that black or blue pens are pulled out.
2. To make sure that there is a not-blue pen has been picked, we must pulled out 19 of the total pens. If so, there must be at least 1 not-blue pen.
3. Assume (the worst possibility) that there is still 12 black pens inside the drawer. Add 19 with 12 to make sure that we have picked at least 1 pen which is not blue or black. This gives us $\boxed {31}$

18 blue 12 black 17 RED

You can pull 30 (18 + 12) in a row and still end up without a red pen. Draw one more, and the last one is guaranteed to be a red pen. 31 is the answer

SHE MIGHT BE TOO UNLUCKY TO GET A RED PEN UNTIL ALL THE BLUE AND RED PENS ARE TAKEN OUT - therefore total no. of pens she need ton take out to get atleast 1 red pen is 18+ 12+1 = 31

Add the blue and black pens togheter, as if she would pull out all of them. Then add 1 - wich will surely be a red one. ( the answer is 31).

i agree with Peng Ying Tan amazing job man

IF BELLA GETS A RED PEN THEN SHE NEVER SEARCHES AGAIN FOR PENS . ALSO THERE MAY BE A CHANCE OF PICKING 18 BLUE AND 12 BLACK PENS BEFORE SHE GETS A RED PEN. THEREFORE 18+12+1 = 31

well ! bella have totally 47 pens! if she get 30 pen assume until no red pen come in drawer ..so after that if she get 1 pen (31 pen) that should red pen so the minimum possibility is 31

blue=18+black=12

To make sure that Bella gets a red pen on pulling out the pens, we have to take out all the blue and black pens first so as to make the drawer full of only red pens.

Then, since only one red pen is required, so we have to pull out one more pen from the drawer. So, total no. of pens to be pulled out $=(18+12+1) = \boxed{31}$

If she draws any less than $31$ pens then that may only comprise blue and black pens. As there are in total $30$ blue and black pens, if she draws $31$ pens then at least one has to be red.

Since a red pen is required, sum up all the pens of other colours and add 1. The result is the answer.

given that there are 17 red pens , 18 blue pens and 12 black pens. We may think that at first draw only we can get a red pen, so min. no. of draws is 1, we think. But we can't sure to that the drawn pen is red. If we have drawn all black pens and blue pens we can surely say that next drawn pen is red so the min. no. of pens she has to pull out from the drawer to make sure she gets a red pen is no. of black pens + no. of blue pens + next draw(for red pen) = 12+18+1=31

Suppose that Bella is a very unlucky person. In this worst case scenario, Bella pulls out all the pens that are not red before finally getting a red one. Thus, the minimum number of pens she has to pull out is $18+12+1=\boxed{31}$ .

If she pull out 30 pens, so it can be 18 blue pens + 12 black pens <=> she has to pull out one more. So, the result is 31