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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 $\frac{7}{8}$!

user91500
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Niharika
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    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. – ShreevatsaR Mar 03 '14 at 09:14
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    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… – Riccardo Mar 03 '14 at 09:15
  • There are $2^3=8$ possible outcomes. – AnonSubmitter85 Mar 03 '14 at 09:16
  • Also think that in your line of though if you have 3 childs the first one have a probability of $\frac{3}{4}$ of being a boy. – durum Mar 03 '14 at 11:51
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    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. – gerrit Mar 03 '14 at 15:45
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    @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. – Andrew Coonce Mar 03 '14 at 16:44
  • @gerrit Not so. "Analysis of the effect of multiple birth, birth order, age of parents and the sexes of preceding siblings on the secondary sex ratio was performed for 815,891 children, born in Denmark, 1980–1993. The proportion of males was analysed [...] no independent effect was observed for maternal age, birth order, the sex of the preceding child, or the combination of sexes of previously born children in the family." Jacobsen et al., "Natural variation in the human sex ratio", Hum. Reprod. (1999) 14 (12): 3120-3125. (online) – David Richerby Mar 03 '14 at 19:38
  • @DavidRicherby No independent effect. But P(sex of 3rd child|first 2 children are male) is not necessarily an independent "experiment". In other words, perhaps a family of all males is indicative of a different, known cause that the experimenters in the cited study have controlled for, whereas this question has not. – gerrit Mar 03 '14 at 20:17
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    @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? – David Richerby Mar 03 '14 at 20:41
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    @DavidRicherby I concede that my claim was too strong (my source was my high-school biology teacher, who had a medical explanation that sounded solid to me). I claimed that "A mother who has given birth to 2 boys is statistically more likely...". Correct would have been to replace is by might be, so that for the question to be complete, the assumption should be made that the chances are fully independent. To investigate whether there is any peer-reviewed evidence for my original claim is beyond the scope of this question (and, for that matter, this site). – gerrit Mar 03 '14 at 22:03
  • Ah oh; nobody look - http://en.wikipedia.org/wiki/Human_sex_ratio#Gender_imbalance. Biology be playin' PING PONG with probability. – Question3CPO Mar 04 '14 at 00:15
  • @AndrewCoonce in your lottery example, isn't 1/2 the best answer one can give in the absence of any extra information? – user13107 Mar 04 '14 at 06:44
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    What about Justin Bieber? – evil999man Mar 04 '14 at 14:31
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    Whenever you come across $P($At least one$)$, it is almost always mathematically easier to convert the problem into $1-P($none$)$, as it reduces the problem into finding one probability total! – 2c2c Mar 04 '14 at 11:36
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    I guess we're removing from consideration as to WHY they had 3 children, and not 1 or 2. In many countries, even ours, there is a bias towards WANTING boys, so it may be that if you have 3 you were trying for a boy, which would mean that there is more likely chance that the first two were girls – TruthOf42 Mar 04 '14 at 20:30
  • @TruthOf42 ours? – Joshua Taylor Mar 04 '14 at 21:06
  • @JoshuaTaylor: so much for not making an ass out of myself... though maybe I was trying to convey that we all have a belief that WE aren't bias – TruthOf42 Mar 04 '14 at 21:09
  • Can anyone explain a bit more what EXACTLY is wrong with the first logic, shown in the question? I get that the order of birth is important for all possible combinations, but the first approach seems to work as well on the fight sight. Some in-depth info? – Lazar Ljubenović Mar 05 '14 at 20:02
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    Aaah, now I get it. BBB and BBG are not EQUALLY LIKELY. Since the probability of the birth of 2 Boys and 1 girl can occur in three different ways, i.e, BBG,BGB and GBB but the probability of all three boys can occur in only one way and that is BBB. So if I use my first logic, I am treating UNEQUALLY LIKELY events as EQUALLY LIKELY which is most definitely wrong!!! So the answer should be 7/8. – Niharika Mar 29 '14 at 15:53

6 Answers6

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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.

lab bhattacharjee
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    It is usually easy to think in terms of the complement in these type of problems. Nice answer. – Bennett Gardiner Mar 03 '14 at 10:05
  • So you are saying the order does matter right? – gideon Mar 04 '14 at 08:16
  • @gideon, have you noticed anything like order in the answer? – lab bhattacharjee Mar 04 '14 at 08:46
  • 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) – gideon Mar 04 '14 at 08:49
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    @gideon, no, order is immaterial here. I don't think DonAntonio has meant any order, he has listed the combinations – lab bhattacharjee Mar 04 '14 at 08:51
  • @gideon: The probability of all three children being girls is $1/8$. Now, if you actually care about all possible outcomes of three children, there are two ways of saying it: (1) that there are eight possible outcomes each with equal probability $1/8$, OR (2) that there are four possible outcomes: three girls or three boys, each with probability 1/8, and two-girls-and-a-boy or two-boys-and-a-girl, each with probability $3/8$. The question is not about whether "order matters", the question is how to pick the weights for your outcomes such that the weights match the model you have of the world. – ShreevatsaR Mar 04 '14 at 13:04
  • About to reach a century! – Archer Oct 25 '18 at 17:56
  • Finding the probability of the COMPLEMENT of your event is a powerful tool in probability theory. Learn to use it. – richard1941 Apr 03 '19 at 03:32
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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 $\;\dfrac18\;$ .

DonAntonio
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    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. – DMM Mar 03 '14 at 11:49
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    @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. – DonAntonio Mar 03 '14 at 12:11
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    @DMM you could just as easily say that the probability is $50%$ and it there was an experimental error of $2.6%$ ;) – Guy Mar 03 '14 at 18:47
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    @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? – Nick Matteo Mar 03 '14 at 20:25
  • @Sabyasachi It's actually well established that male births are statistically slightly more likely than female ones (I remember about 104:100 discarding things such as killing female infants). Males are also statistically more likely to die young (more likely to take risks, but also just more likely to do from health complications apparently), so the ratio actually balances itself over time. Interesting fact, although for the actual problem obviously not really important ;) – Voo Mar 03 '14 at 23:35
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    @Kundor why? Physicists are allowed to talk about spherical cows in vacuum and we can't take a 50:50 ratio? – Guy Mar 03 '14 at 23:38
  • same for you @Voo – Guy Mar 03 '14 at 23:39
  • @Sabysachi I'm not saying we should use 104/200 as the probability, because goodness there are several other biological things factoring in too and in the end it's just a nice simple stat homework to get people started with the basics. It's just an interesting tidbit which I personally find quite enjoyable to just mention on the side x) – Voo Mar 03 '14 at 23:42
  • @DonAntonio even what you mean is that the order of the births matter right? – gideon Mar 04 '14 at 08:20
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    @Sabyasachi reminds me of my favourite joke from The Big Bang Theory. "(story about asking a physicist for help with his chickens that won't lay eggs) Okay, I have a solution, but it'll only work for spherical chickens in a vacuum!" – Cruncher Mar 04 '14 at 21:32
  • @gideon kind of. Think about it intuitively. If you want exactly 1 boy out of 3, after the first child is born, regardless of gender, it's still possible to get your 1 out of 3 based on the rest of the births. If you want exactly 3 boys, if you get born a girl, you're done already. Intuitively this should make perfect sense that "order matters". – Cruncher Mar 04 '14 at 21:35
  • @Cruncher actually my joke was a reference to that particular big bang theory episode. It's a little weird that that joke was intended to be not understood, so that the audience could laugh at Leonard's nerdiness/stupid jokes, but so many people actually got it. It cracked me up. :D – Guy Mar 05 '14 at 09:38
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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).

TooTone
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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.

-2

The possibilities are

ggg ggb gbg bgg gbb bgb bbg bbb

at least one boy... 7/8

Mark Williams
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    this is a duplicate of an existing answer http://math.stackexchange.com/a/697515/77151. Did you miss it? – TooTone Mar 04 '14 at 18:42
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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
user23715
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  • 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. – Larry Gritz Apr 16 '18 at 20:10
  • @LarryGritz - If true then the question belongs in the Meta portion of SE. But thanks for noticing my attention to detail :) – user23715 Jul 04 '18 at 03:51