what is the probability that we get a blue blue and a red ball with three tries From all the balls in the urn $0.4$  are blue and $0.25$ are red. What is the probability that after picking 3 balls we get a blue and a red ball.
My reasoning is as follows. All the different ways we can pick the balls are $3!$ so we have multiply this by the probability of getting one blue and one red.
$$3!0.4*0.25 = 0.6$$
But I'm not sure because I ran few rows of excel (what I think was similar simulation for this task... not sure) and i got answer around $0.4$. 
Please tell me if I'm right and if I'm wrong all the tips are welcome. Thank you :)
 A: To elaborate on the solution sketched in the comments:
First of all, I assume that the total number of balls is large enough to ensure that the probabilities do not change significantly as we draw some.  If you don't wish to make this assumption, then you need to give us some more information regarding the number of balls.  
Note:  to avoid the problem you could say that you draw "with replacement".  That certainly ensures that the probabilities do not change on subsequent draws.
Second, I will assume that you are after "at least one red and at least one blue." so that, for example, getting two reds and a blue would be a success.  This is consistent with your simulation result (again, if you intended something else, you should say so).
Given those assumptions,  there are three "good" scenarios:  two reds and a blue, two blues and a red, one red one blue and one other.  We'll do each in turn:
Scenario I:  two reds and a blue.  If we fix the order, say $RRB$ then the probability is $.25\times .25\times .4$.  As there are three possible orders, the probability here is $3\times .25\times .25\times .4=\boxed {.075}$
Scenario II:  two blues and a red.  Similar.  We get $3\times .25\times .4\times .4=\boxed {.12}$
Scenario III:  one red one blue one other.  If we fix the order, say $RBX$, the probability is $.25\times.4\times .35$.  As there are $3!=6$ possible orders the total probability here is $6\times .25\times.4\times .35=\boxed {.21}$
The final answer (depending on the two assumptions spelled out at the start) is then the sum $$.075+.12+.21=\boxed {.405}$$
A: I preassume that the probabilities do not change relevantly by taking out two balls.
Let $B$ denote the event that one of the drawn balls is blue.
Let $R$ denote the event that one of the drawn balls is red.
Then: $$P(B\cap R)=1-P(B^{\complement}\cup R^{\complement})=$$$$1-P(B^{\complement})-P(R^{\complement})+P(B^{\complement}\cap R^{\complement})=1-0.6^3-0.75^3+0.35^3=0.405$$
A: 
as pointed in the comments I forgot to exclude the scenarios of RRR and BBB, RRW, BBW. So the following solution is wrong.

I'll assume that the probability of picking a red ball and a blue ball means the probability of picking at least one red and one blue. So
$$P(\textrm{1 red and 1 blue})= P(\textrm{1 white $\cup$ 0 whites})=P(\textrm{1 white})+P(\textrm{0 white})$$
$$=\frac{{white \choose 1}{blue+red \choose 2}}{white+red+blue \choose3}+\frac{blue+red \choose 3}{white+red+blue \choose 3}$$
Now, suppose there are 20 balls, according to the specifications, 8 are blue, 5 are red and 7 are white, so 
$$=\frac{{7 \choose 1}{13 \choose 2}}{20 \choose3}+\frac{13 \choose 3}{20 \choose 3} = 0.7298246 $$
Now, if there are 100 balls, 40 are blue, 25 are red and 35 are white, so
$$=\frac{{35 \choose 1}{65 \choose 2}}{100 \choose3}+\frac{65 \choose 3}{100 \choose 3} = 0.7203463$$
so, you see, the probability is different, depending on the total amount of balls. Are you sure all the information of the question has been given?

side note: If you're not used to $a \choose b$ it means the amount of subsets of size $b$ in a set of size $a$ or "how many ways you can pick $b$ balls in an urn with $a$ balls", ${a \choose b }=\frac{a!}{(a-b)!b!}$
