# Assume that the probability is $\frac12$ that a child born is a boy

Assume that the probability is 1/2 that a child born is a boy. If a family has 3 kids. What is the probability that they have:

a) exactly one boy? $3/8$ is that right?

b) at most two girls? $7/8$ is that right?

There are only $2^3 = 8$ possible ways to arrange the genders boy/girl, with repetition. Since the probability of boy and girl are equal, the probability of each of these arrangements are also equal. We can therefore count the number of possible "good" answers, and divide by 8.

There are 3 possible ways to have 1 boy, that being "first/second/third child is boy", so the probability of exactly 1 boy is indeed $\frac{3}{8}$.

Having at most 2 girls is the opposite of having three girls, so since there are only one way of having three girls, there must be $8 - 1 = 7$ ways of having at most two girls, so the probability of at most two girls is indeed $\frac{7}{8}$

Assume $X_i$ = gender of ith kid , $B$ = Boy, $G$ = Girl

$$P(X=B) = P(X=G) = \frac12$$

If we assume birth of kids are independent of each other, then : \begin{align} P(\text{exactly one boy}) &= P(X_1=B)P(X_2=G)P(X_3=G)\\ &+\space P(X_1=G)P(X_2=B)P(X_3=G)\\ &+\space P(X_1=G)P(X_2=G)P(X_3=B)\\ &= 3(1/2)^3 \\ & = 3/8 \end{align}

\begin{align} P(\text{at most two girls}) &= 1 - P(\text{three girls})\\ & = 1 - P(X_1=G)P(X_2=G)P(X_3=G) \\ &= 1 - (1/2)^3 \\ &=7/8 \end{align}

• omg what's that Commented Aug 7, 2013 at 7:23
• This answer is with the hypothesis of gender of kids being independent of each other. Commented Aug 7, 2013 at 7:24
• So what are you saying the answer is ? I can't understand that sorry. Commented Aug 7, 2013 at 7:25
• Do the formulas in the arguments, namely $P(X=B)$ and $P(B=1)$, speak about different things? First you say $X$ is the gender, but then you count occurences. Commented Aug 7, 2013 at 7:36
• @NickKidman Now it would be how you like! Commented Aug 7, 2013 at 7:41

Let $G$ be the event that a girl is born and $B$ that a boy is born. All in all we pick 3 times from the set $\{G, B\}$. That gives us $2^3 = 8$ possibilities for girl/boy combinations (e.g. GGB, GBG, BBB, etc.).

a) Exactly one boy is all combinations with just one B, that's BGG, GBG, GGB. Therefore in $P(\textrm{exactly one boy}) = 3/2^3 = 3/8 \approx 0.375$ cases we have a boy.

b) At most two girls is either 0 girls, 1 girl or 2 girls, that's the case for all combinations except for GGG. Therefore $P(\textrm{at most two girls}) = 7/8 =0.875$.

If we let X=the number of boys, then X is distributed with a Binomial distribution n=3, p=1/2.

Probability of exactly one boy: P(X=1) = 3C1*(1/2)^1*(1/2)^2 = 3/8

Probablity of at most two girls = Probability of at least one boy: P(X>=1) = 1-P(X=0) = 1-(3C0*(1/2)^0*(1/2)^3) = 1-1/8 = 7/8