36
$\begingroup$

According to me there are $4$ possible outcomes:

$$GGG \ \ BBB \ \ BGG \ \ BBG $$

Out of these four outcomes, $3$ are favorable. So the probability should be $\frac{3}{4}$.

But should you take into account the order of their birth? Because in that case it would be $\frac{7}{8}$!

$\endgroup$
19
  • 23
    $\begingroup$ BBG and BGG are each three times as likely as BBB or GGG. See this question and the comments: for probability, you need to also assign weights to the outcomes, not just count them. $\endgroup$
    – ShreevatsaR
    Commented Mar 3, 2014 at 9:14
  • 2
    $\begingroup$ For counting problems like this it is better to denote the three child as $A,B,C$ and counting each case of the form $A$ is male , $B$ is female, $C$ is female and so on… $\endgroup$
    – Riccardo
    Commented Mar 3, 2014 at 9:15
  • 6
    $\begingroup$ A mother who has given birth to 2 boys is statistically more likely to give birth to another boy, than to a girl. It seems the answers all seem to assume that P(B)=P(G)=½. This assumption is only approximately correct. $\endgroup$
    – gerrit
    Commented Mar 3, 2014 at 15:45
  • 8
    $\begingroup$ @Niharika: To understand why your logic is incorrect, imagine applying it to purchasing a lottery ticket: there are two outcomes and one is favorable, so your odds of winning would be 1/2. This is intuitively not true for the same reason your logic is not. $\endgroup$
    – alc
    Commented Mar 3, 2014 at 16:44
  • 4
    $\begingroup$ @gerrit Can you cite any research that supports your claim that a mother who has given birth to two boys is statistically more likely to give birth to another boy? $\endgroup$
    – David Richerby
    Commented Mar 3, 2014 at 20:41

6 Answers 6

Reset to default
115
$\begingroup$

The complement of at least one boy is all three girls

So, $P($ at least one boy$)=1-P(GGG)$

$=\displaystyle1-\left(\frac12\right)^3$

This is the de facto way of solving problems of Probability of at least one in case of Binomial Distribution like tossing a coin etc.

$\endgroup$
8
  • 6
    $\begingroup$ It is usually easy to think in terms of the complement in these type of problems. Nice answer. $\endgroup$
    – Bennett Gardiner
    Commented Mar 3, 2014 at 10:05
  • $\begingroup$ So you are saying the order does matter right? $\endgroup$
    – gideon
    Commented Mar 4, 2014 at 8:16
  • $\begingroup$ @gideon, have you noticed anything like order in the answer? $\endgroup$
    – lab bhattacharjee
    Commented Mar 4, 2014 at 8:46
  • $\begingroup$ I'm asking as a newbie :) @labbhattacharjee your answer suggests P(GGG) is 1/8 (which it is if there are possibilities, which will have to include order like DonAntonios answer) $\endgroup$
    – gideon
    Commented Mar 4, 2014 at 8:49
  • 1
    $\begingroup$ @gideon, no, order is immaterial here. I don't think DonAntonio has meant any order, he has listed the combinations $\endgroup$
    – lab bhattacharjee
    Commented Mar 4, 2014 at 8:51
41
$\begingroup$

There are in fact eight possible outcomes:

$$GGG\,,\,GGB\,,\,GBG\,,\,BGG\,,\,BBB\,,\,BBG\,,\,BGB\,,\,GBB$$

Of these, only one does not include a boy (B) in the event, and thus the probability of all girls is $\;\dfrac18\;$ .

$\endgroup$
12
  • 7
    $\begingroup$ This assumes, of course, that the probability of male and female births is the same. The 2012 (most recent available) data for England and Wales published by the UK's Office for National Statistics shows a bias towards male births (51.3% of live births). I would expect the statistics for other regions to differ slightly from 50:50. $\endgroup$
    – DMM
    Commented Mar 3, 2014 at 11:49
  • 13
    $\begingroup$ @DMM Yes, of course. It could be that upon checking one single casino and a very fair, well-balanced die, the probability to get some number (say, 3) is slightly higher than $\;1/6\;$ (because of the hand throwing the dice, the table, the material the die is made of, etc.), but we usually assume some events to have "the usual, standard" probability to happen. $\endgroup$
    – DonAntonio
    Commented Mar 3, 2014 at 12:11
  • 1
    $\begingroup$ @DMM you could just as easily say that the probability is $50\%$ and it there was an experimental error of $2.6\%$ ;) $\endgroup$
    – Guy
    Commented Mar 3, 2014 at 18:47
  • 2
    $\begingroup$ @DonAntonio Except that every measurement everywhere shows that more males are born than females. Yes, you can arbitrarily declare some ratio the "standard" probability, but why? $\endgroup$
    – Nick Matteo
    Commented Mar 3, 2014 at 20:25
  • 1
    $\begingroup$ @Kundor why? Physicists are allowed to talk about spherical cows in vacuum and we can't take a 50:50 ratio? $\endgroup$
    – Guy
    Commented Mar 3, 2014 at 23:38
24
$\begingroup$

Another way to look at this is to draw this out

enter image description here

Here I follow the stereotypical association of gender and colors: the blue boxes represent boys and the pink boxes represent girls. Each time you have a boy or a girl, in the next generation you can have a boy or a girl also, so the number of possibilities is doubled each generation.

In terms of your problem, when you have a boy, that represents a checkmark against "at least one of them is a boy", so I've crossed the box concerned. However all the subsequent generations after this boy are also families in which there is at least one boy, so I've crossed those out too. You can see that the chance of having at least one boy is $1/2$ in the first generation, $3/4$ in the second, and $7/8$ in the third. This generalizes to $(2^n-1)/2^n$ in the nth generation.

Conversely the chance of having no boys is $1/2$ in the first generation, $1/4$ in the second, and $1/8$ in the third. This generalizes to $1/2^n$ in the nth generation.

(Essentially I've drawn a probability tree diagram here, which generalizes to much more complicated problems).

$\endgroup$
2
  • $\begingroup$ Should your second-last sentence be "the chance of not having at least one boy"? $\endgroup$
    – Andrew Leach
    Commented Mar 4, 2014 at 18:37
  • $\begingroup$ @AndrewLeach thankyou, brain dzzzzt $\endgroup$
    – TooTone
    Commented Mar 4, 2014 at 18:39
3
$\begingroup$

Assuming that the probability of getting a transgender=0 we have the probability of getting no boy=1/8,because it can only be by ggg, where g represents a girl child.So,the required result is the complement of the above mentioned event whose probability is clearly 7/8.

$\endgroup$
-2
$\begingroup$

The possibilities are

ggg ggb gbg bgg gbb bgb bbg bbb

at least one boy... 7/8

$\endgroup$
1
-2
$\begingroup$

I’m adding an answer since I’m not reputable enough to add a comment to “lab bhattacharjee”’s answer. There are some assumptions to consider here:

Assuming that the family is generated the old fashioned way then “lab bhattacharjee”’s answer is mostly correct with only a few more explicit assumptions, as follows.

  1. Assuming that the slight natural skew towards human male progeny is ignored
  2. Assuming adoption is not under consideration
  3. Assuming modern medical reproductive techniques are not being used, including...
  4. Assuming no selective abortion, then only...
  5. Assuming we are counting the children and not the parents. Because given that the father is male, that makes the “real” answer to this question, as given by “Niharika” and edited by “Ric Ped”, out to be 100% and no statistics are needed
$\endgroup$
2
  • $\begingroup$ It should be obvious from context (including what information is not given) that the original question is an exercise in reasoning about how to compute probabilities, not about human families in the real world, and therefore this degree of getting lost in the weeds does not actually address the original question at all. $\endgroup$
    – Larry Gritz
    Commented Apr 16, 2018 at 20:10
  • $\begingroup$ @LarryGritz - If true then the question belongs in the Meta portion of SE. But thanks for noticing my attention to detail :) $\endgroup$
    – user23715
    Commented Jul 4, 2018 at 3:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .