0

Let a family have two children. It is known that one of the children is a boy. What is the probability that both the children are boys.

So for this we build the sample space $S=\{(b,b)(b,g)(g,g)\}$

Let our event E be the case where both children are boys

$E=\{ (b,b) \}$

Let the conditional be F

$F=\{(b,g),(b,b)\}$

Hence $P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{1/3}{2/3}=\frac{1}{2}$

But the answer in my book is given as $\frac{1}{3}$ and I can't seem to understand why.

RobPratt
  • 45,619
Boy
  • 427

2 Answers2

2

Order matters, so your sample space should be $S=\{(b,b),(g,g),(b,g),(g,b)\}$. For example if we have $(g,b)$, then a girl was the first child and a boy was the second child.

Now, given that one of the children is a boy, our remaining possibilities are $$(b,b),(b,g),(g,b).$$ Among these three options, only $(b,b)$ corresponds to the other child being a boy as well. Since there's only one favorable outcome out of the three possible outcomes, the probability you want is $1/3$, which agrees with what your book says.

ilovebulbasaur
  • 1,724
  • But the text of the question has nothing to say about which order they were born in. For example if were to select two balls from two choices of colours, the possible ways to do that is 3C2=3. So similarly with the boy and girl case the order should not matter but it seems like it does. Can you explain more in detail why it does? – Boy Aug 28 '20 at 03:46
  • If we were only concerned about the possible genders but not the order which they occur, then there are only three ways - BG, GG, BB. But we need to take into account not only the possible outcomes, but also the fact that each outcome has a different probability. In this case, GG and BB each occur with probability 1/4, while GB occurs with probability 1/2. My point is, you could solve the problem using a sample space with unordered outcomes, except you would have to assign the probabilities to the unordered outcomes correctly. – ilovebulbasaur Aug 28 '20 at 07:35
  • So, your last line should have been $\mathbb{P}(E|F)=\frac{(1/4)}{(3/4)}=1/3.$ – ilovebulbasaur Aug 28 '20 at 07:37
  • Thank you for that explanation but I still do not understand why GB and BG occur as two equiprobable separate events. If you were to consider the probability of a family having two kids not of the same sex, would it be 1/2 or 1/3? – Boy Aug 28 '20 at 13:01
  • 1
    Think about it as a coin toss - associate "boy" to "heads" and "girl" to "tails" (no sex discrimination intended). Then the probability of tossing two coins in a row, and getting two coins (kids) not of the same side (sex) is $1/2$. – ilovebulbasaur Aug 28 '20 at 13:50
  • 1
    @Boy Imagine you have $100$ coins which at first glance all appear identical. You somehow pick them all up in a bag, mix them up, and dump the contents of a bag onto a table and count the number of heads that show. Yes, you will only be able to discern $101$ different possible outcomes... the outcome with zero heads, with $1$ head and $99$ tails, with $2$ heads and $98$ tails, etc... Despite there only being $101$ discernible outcomes, that does not imply the probability of each is $\frac{1}{101}$. Suppose your friend who is more knowledgeable in coins comes along and points out... – JMoravitz Aug 28 '20 at 15:55
  • 1
    ...that each coin was made in a different year. Your friend has the ability then to look at these coins and be a lot more precise in how he writes out the outcomes, writing down whether the coin from the year 1900 landed head or tail, whether the coin from the year 1901 landed head or tail, etc... When comparing his sample space to yours, both are equally correct and valid sample spaces to use when talking about counting how many coins landed heads. However... here's the kicker... his will have the outcomes be equally likely. This is because of how we model the scenario mathematically... – JMoravitz Aug 28 '20 at 15:58
  • 1
    ...The only reasonable way to model this is by assuming each coin, being a distinct physical entity, whether distinguishable by our sight or not from one another, will have a chance of either outcome independently of the results of any other coin. You will find the probability of all $100$ coins being heads is tremendously low (only $\frac{1}{2^{100}}$), far lower than an even split. The same is true for the scenario of children and genders. While as an observer we might not know which child is older/younger... the fact remains one child is (even if only by a few seconds for twins). – JMoravitz Aug 28 '20 at 16:00
  • @JMoravitz I understand it very well now. Your explanation had been very helpful. – Boy Aug 29 '20 at 11:35
2

The events in your sample space are not equiprobable. $(b,g)$ has probability $\frac12$, because the oldest child might be a boy or a girl. Alternatively, and perhaps more clearly, add the event $(g,b)$ to the sample space and make all events equiprobable.

saulspatz
  • 53,131
  • But the text of the question has nothing to say about which order they were born in. For example if were to select two balls from two choices of colours, the possible ways to do that is 3C2=3. So similarly with the boy and girl case the order should not matter but it seems like it does. Can you explain more in detail why it does? – Boy Aug 28 '20 at 03:43
  • 1
    @Boy This case is more akin to flipping coins than extracting balls from a barrel. The probability for two boys $1/4$, for two girls is $1/4$, leaving the probability for one from each as $1/2$. That is: $$\dfrac{{^2\mathrm C_2}{^2\mathrm C_0}}{2^2},~\dfrac{{^2\mathrm C_0}{^2\mathrm C_2}}{2^2},\text{ and }\dfrac{{^2\mathrm C_1}{^2\mathrm C_1}}{2^2}$$ – Graham Kemp Aug 28 '20 at 04:01