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.
Best Answer
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*}