What is the probability that in a family of 5 children, girls outnumber boys. (Assuming probability of a boy is ${\frac{1}{2}}$)

Question

How I look at this is that the total number of outcomes is $5!$ and in the given event, either there are $3,4$ or $5$ girls in the family. Now I should calculate the total number of possible outcomes where there are 3 girls and 4 girls and 5 girls and hence find the probability. But is this approach correct? Also is there any shorter approach to this question?

Answer 1

The total number of families is $2^5 =32$: from $GGGGG$ to $BBBBB$, we have an sequence of $5$ children and for each child we can have a boy or a girl. All of these sequences are equally likely.

How many sequences have no boys, 1 boy or 2 boys?

• no boys: just one ($GGGGG$)
• one boy: just $5$: the one boy can be the first, second, etc. till the fifth.
• two boys: just $10$: the two boys an be at $\binom{5}{2}$ places: we choose the two positions out of the $5$ without order and no replacement.

So in $16 = 1+5+10$ of the outcomes, girls outnumber boys. So the probability is $\frac{1}{2}$. This makes sense in hindsight: there can be no "draw": either there are more boys or more girls. And by symmetry, these must have the same probability.

Answer 2

I would not bring the number $5!$ into this problem. What do you think the $5!$ outcomes would be?

You can assume the children are born in a some sequence, that is, there is an oldest child, then a second oldest, and so forth. The oldest child can be a boy or girl, the second oldest can be a boy or girl, and so forth. You get the choice "boy or girl" five times, so you could count outcomes the same as you count the outcomes of tossing a coin five times.

But the really simple approach to this particular problem is:

Girls outnumber boys in the following circumstances: 3 girls, 4 girls, or 5 girls.

Boys outnumber girls in the following circumstances: 3 boys, 4 boys, or 5 boys.

Given that for each birth the probability of a boy is the same as the probability of a girl (both are $\frac12$), which is more likely: 3 girls or 3 boys?

Ask the same question for the "4" and "5" cases and see whether it is more likely for girls to outnumber boys, or boys to outnumber girls. Also note that since the number of children is odd, it is impossible to have the same number of boys and girls; one must outnumber the other.

Answer 3

You have to create a tree diagram to get all possible sample space. then see how many space have girls outnumbered. I can't draw it here. There are 32 possible outcomes. And 16 outcomes have no of girls 3 or more. So probability of outnumbered girls is 16/32 = 1/2

The possible outcomes are like these BBBBB, BBBBG, BBBGB, BBBGG, BBGBB,BBGBG,BBGGB,BBGGG, ---------------GGGGG

Answer 4

If for a random chosen family with $5$ children $G$ denotes the number of girls and $B$ denotes the number of boys, then - if the probability on a boy equals the probability of a girl - $(G,B)$ and $(B,G)$ have the same distribution.

This implies that: $$P(G>B)=P(B>G)$$

Next to that we have: $$P(G>B)+P(B>G)=1$$

Now draw conclusions.

Answer 5

If you assume that the gender of each child born is uniformly distributed and is independent of the gender of each other child... this becomes identical to the question of "What is the probability that you flip more heads than tails in five consecutive coin tosses."

There will be $2^5=32$ equally likely possible outcomes: $\{HHHHH, HHHHT,HHHTH, HHHTT, HHTHH, HHTHT,\dots, TTTTH, TTTTT\}$. If you really wanted to you could finish writing them all out.

Now... as for the probability that in our five flips there are strictly more heads than tails, we could highlight each of the above outcomes that correspond to that and count with our fingers and toes and then divide the result by $32$ to get the probability... but the whole point of an introductory course on combinatorics or probability is to avoid having to count by using brute force whenever possible!

Instead, we could count faster by noting that there is exactly one way to have all five of the flips as heads, five ways to have all of the flips but one as heads, and ten ways to have all but two of the flips as heads. That is, $\binom{5}{0},\binom{5}{1},$ and $\binom{5}{2}$ respectively. Adding these together we get $\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16$ so the total number of ways of flipping strictly more heads than tails in five consecutive flips of a fair coin as $\frac{16}{32}$.

We could have arrived at this using a well known formula instead without having to rederive it on the spot by utilizing the binomial distribution, giving us the probability of more heads than tails as $\sum\limits_{k=3}^5\binom{5}{k}(\frac{1}{2})^k(\frac{1}{2})^{5-k}$ which again evaluates as $\frac{1}{2}$.

Both of the above required some knowledge of how to find the probability of an exact number of heads versus tails... but even that isn't necessary as already pointed out in other answers and comments.

Note that exactly one of the following three situations will always occur: There are more heads than tails, There are more tails than heads, There are the same number of heads and tails.

This implies: $$1 = Pr(\text{more heads than tails})+Pr(\text{more tails than heads}) + Pr(\text{same amount})$$

Since we flipped an odd number of coins, $Pr(\text{same amount})=0$. Then, noticing that $Pr(\text{more heads than tails})$ should equal $Pr(\text{more tails than heads})$ due to symmetry, we learn that $1=2Pr(\text{more heads than tails})$ implying the probability of more heads than tails is simply $\frac{1}{2}$.

Of course, all this talk about coin flipping is perfectly analogous to the question about genders of children (given the obvious assumptions of uniformity and independence of the randomness of the gender of each child).

Answer 6

Your very first instinct should be that boys out number girls is the same probability of girls out numbering boys. Your second instinct should be $5$ is odd so either boys outnumber girls or girls outnumber boys and as they are both equally probable, the probability is $\frac 12$.

Now, all that remains is justifying our instincts.

For notation purposes: Let $(a,b,c,d,e)$ be an instance of the five family members where each $a,b,c,d,e$ is either $B$, a boy, or $G$ a girl. Will notate $\overline x$ to mean if $x=B$ then $\overline x = G$ and if $x = G$ then $\overline x= B$.

So for every instance $(a,b,c,d,e)$ there is an exact opposite $(\overline{a,b,c,d,e})$. So if $(a,b,c,d,e)$ is a case where boys outnumber girls then $(\overline{a,b,c,d,e})$ is a case where girls outnumber boys. And vice versa.

So the cases where girls outnumber boys are in one-to-one corespondence with the cases where the boys outnumber the girls. i.e. the number of cases of one is the same as the number of cases of the other.

So $P(\text{boys outnumber girls})=\frac{\text{# boys outnumber girls}}{\text{# total cases}}=\frac{\text{# girls outnumber boys}}{\text{# total cases}}=P(\text{girls outnumber boys})$.

So our first instinct is justified.

To have neither girls nor boys outnumber each other is to have and equal number of boys and girls (if we are assuming strict binary gender which we really should not) and as $5$ is odd this is impossible.

So our second instinct is correct.

So as boys outnumbering girls, girls outnumber boys, and girls equaling boys are mutually exclusive and exhaustive we have:

$1 = P(\text{boys outnumber girls}) + P(\text{girls outnumber boys})+P(\text{girls equal boys})$

$=P(\text{boys outnumber girls}) + P(\text{girls outnumber boys})+0$

$= P(\text{girls outnumber boys}) + P(\text{girls outnumber boys})$

$= 2P(\text{girls outnumber boyss})$

so

$P(\text{girls outnumber boys}) = \frac 12$