In a family with 3 children, what is the probability that they have 2 boys and 1 girl? I'm doing some regents practice questions, and one of them asked

In a family with 3 children, what is the probability that they have 2
  boys and 1 girl?

And the answer choices are 


*

*3/8 

*1/4

*1/8

*1/2


My teacher said the answer was choice one but I'm having trouble understanding why.
My approach was to draw out the probabilities, since we have 3 children, and we are looking for 2 boys and 1 girl, the probabilities can be Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy. So a 2/3 chance, but I don't get how it's a 3/8 chance. Any help is appreciated.
 A: No, the possible outcomes are
$$\mathsf{BBB, \boxed{\mathsf{BBG}}, \boxed{\mathsf{BGB}}, \boxed{\mathsf{GBB}}, BGG,\mathsf{GBG}, GGB, GGG}$$
wherein $3$ meet the requirement. There are $8$ possible outcomes, all equally likely (if we assume each gender is equally likely).  Hence the choice is $3/8$.
We can also think about it in at least one more way: 
You identified all the possible ways to get 2 boys and one girl. Since the events are disjoint, we can add up the probabilities
\begin{align*}
P(\text{2 boys, 1 girl}) &= P(\mathsf{BBG})+P(\mathsf{BGB})+P(\mathsf{GBB}) \\
&= \frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2} \\
&= \frac{3}{8}.
\end{align*}
A: Since there are two genders and three children, that's $2^3=8$ permutations. You had three permutations. They would be BBB, BBG, BGB, BGG, GBB, GBG, GGB, and GGG. Out of those, the permutations with two boys and one girl are BBG, BGB, and GBB. The answer is the ratio of how many choices fit your conditions to the total number of possibilities, $\frac{3}{8}$.
A: A systematic way to calculate @probablyme s answer to encode the letters B (boy) and G (girl) as a 3 digit binary number with the logical value 1 for true and 0 for false ( that the digit represents a boy ). 
Then we loop through all combinations, for example by incrementing by 1  each step.$$000\to001\to010\to011\to100\to 101\to110\to111$$
We then count the total number of "ones" equals 2. This is the same as the "population count" function used in this question and answer.
The output of the function would be:
$$0\to1\to1\to2\to1\to2\to2\to3$$
And as we can see, for 3 of the 8 states the function evaluates to 2. This means that after counting all possibilities, we found that 3 of the 8 equiprobable states correspond to two boys. This gives a probability of $\frac{3}{8}$.

The strength of this approach is of course that if we modify the question to account for more advanced things, these can be included with fast logical bit-wise operations on the numbers!
A: I drew it out this way:

So the answer is $\frac{3}{8}$.
Another way to think about it. After a first boy we need one of each (50%). After a first girl we must have only boys (25%). So the probability is $.5*\frac{1}{2} + .25*\frac{1}{2} = .375$ which is also equal to $\frac{3}{8}$.
A: Since the question does not mention the order, the possible combinations are 


*

*3 Boys

*2 Boys and 1 Girl (3 ways to do it)

*1 Boy and 2 Girls (3 ways to do it)

*3 Girls


That means the correct answer is 37,5% or 3/8
A: because 'not a boy' is same as a girl (opposite), it could be calculated like this.
$ \binom{3}{2}  (\frac{1}{2})^2(\frac{1}{2})^{2-1} = 0.375 $

$ \binom{n}{k}  (p)^k(1-p)^{n-k} $
A: The following method avoids any need to consider the number of permutations with 2 boys and 1 girl.
3 children must include either at least 2 boys or at least 2 girls (but not both). Hence:
P($\geq 2$ boys) + P($\geq 2$ girls) = $1$.
By symmetry, P($\geq 2$ boys) = P($\geq 2$ girls) = $1/2$.
P(3 boys) = $(1/2)^3 = 1/8$.
Hence P(2 boys, 1 girl) = P($\geq 2$ boys) - P(3 boys) = $1/2 - 1/8 = 3/8$.
