Each bead in an urn is marked with...

Hello LUANDATO. I hope you're well.

This is a tough question, but let's take a look at it.

We only have to take care about the remainders.

* It has been removed 15 numbers whose remainder when divided by 5 is 1.

The product of these 15 numbers could be write as $$R_1=\left(5n_1+1\right)\cdot\cdot\cdot\left(5n_{15}+1\right).$$

* It has been removed 18 numbers whose remainder when divided by 5 is 2.

The product of these 18 numbers could be write as $$R_2=\left(5m_1+2\right)\cdot\cdot\cdot\left(5m_{18}+2\right).$$

* It has been removed 13 numbers whose remainder when divided by 5 is 3.

The product of these 13 numbers could be write as $$R_3=\left(5j_1+3\right)\cdot\cdot\cdot\left(5j_{13}+3\right).$$

* It has been removed 3 numbers whose remainder when divided by 5 is 4.

The product of these 3 numbers could be write as $$R_4=\left(5h_1+4\right)\left(5h_2+4\right)\left(5h_3+4\right).$$

Now, we can see that $$\left(5t_1+r\right)\cdot\cdot\cdot\left(5t_P+r\right)=5W+r^P.$$ So,

$$R_1=5W_1+1^{15}=5W_1+1.$$
$$R_2=5W_2+2^{18}=5W_2+262144=5W_2+262140+4$$ $$=5W_2+5\cdot52428+4=5\left(W_2+52428\right)+4.$$ $$R_3=5W_3+3^{13}=5W_3+1594323=5W_3+1594320+3$$ $$=5W_3+5\cdot318864+3=5\left(W_3+318864\right)+3.$$
$$R_4=5W_4+4^3=5W_4+64=5W_4+60+4$$ $$=5W_4+5\cdot12+4=5\left(W_4+12\right)+4.$$

Finally, the product of ALL numbers is $$T=R_1\cdot R_2\cdot R_3\cdot R_4=5H+1\cdot4\cdot3\cdot4=5H+48=5H+45+3$$ $$=5H+5\cdot9+3=5\left(H+9\right)+3.$$

It implies that the remainder when T is divided by 5 is equal to 3. So, the color of the bead should be YELLOW.

So, the correct answer is D.

I really hope this can help you.

As I said, we are only interested in the remainders.

Feel free to ask me any doubt you could have.