9
$\begingroup$

I know this is a simple problem but I am arguing with a friend about its solution, so I want to show him an "official" proof! Suppose that in any birth, the probability to have a boy is $48.5\%$.

  1. If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?
  2. If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

For the first question, we calculate the probability of one NOT having a boy, which is $1-0.485 = 0.515$ and then the required probability of all three not having a boy is $0.515^3 = 0.1365$ so the probability that at least one will have a boy is $1-0.1365 = 0.8634 = 86.34\%$.

For the second question, since the three events are independent, the probability that all three will have a boy given that at least one will have a boy is equal to the probability that the other two will have a boy. Is it $0.485^2$? I am not sure about the second one.

$\endgroup$
  • $\begingroup$ For $2$ , you just need Bayes' theorem. $\endgroup$ – Peter Dec 5 '20 at 10:23
  • $\begingroup$ @CarlosLopez Technically, some people of all genders; genders, like all abstract concepts, still cannot bear human young. $\endgroup$ – wizzwizz4 Dec 5 '20 at 18:46
  • 4
    $\begingroup$ For 2, is it specified whether you know that a particular person will give birth to a boy or whether you just know that there is at least one boy among the children? This is an important distinction and affects the answer. $\endgroup$ – Magma Dec 5 '20 at 19:19
6
$\begingroup$

Let $X\in\{0,1,2,3\}$ be the number of boys. Let $A$ be the event that $X=3$ and $B$ be the event that $X\ge1$. By definition, we have ${\rm P}(A\cap B)={\rm P}(A\mid B){\rm P}(B)$ so $${\rm P}(X=3\mid X\ge1)=\frac{{\rm P}(X=3\cap X\ge1)}{{\rm P}(X\ge1)}=\frac{{\rm P}(X=3)}{{\rm P}(X\ge1)}.$$ You have calculated ${\rm P}(X\ge1)$ in the first part and evaluating ${\rm P}(X=3)$ is straightforward.

$\endgroup$
6
$\begingroup$

In the second problem, please note your set excludes the case where none of the fetus is a baby boy as you know at least one of them is.

So the probability in the second case that all three fetuses are baby boys $ \displaystyle = \frac{0.485^3}{1-0.515^3}$

$\endgroup$
2
$\begingroup$

now, I am not sure about an "official" sol but here is mine.

now in these cases, I would use sets and a Venn diagram. let's name the persons 1,2,3 now let A = {events|1 has a boy} and similarly for B and C so the Venn diagram would look like-Venn diagram

so now the if at least one of them has a boy then we are in A$\cup$B$\cup$C and so the required probability is n(A$\cap$B$\cap$C)/n(A$\cup$B$\cup$C) = $0.485^3/(1-0.515^3)$

$\endgroup$
1
$\begingroup$

As others have suggested, you can "just" apply Bayes' theorem. However, this problem is a relatively simple one, and so it might help to draw out a probability tree to help organize your thoughts. I will not claim that this is the "best" or most efficient way to approach the problem, but I find that it can help to give some insight into why, as you are relying on something a little more concrete than a statement of a theorem. For this problem, the tree is something like the following:

enter image description here

I have used blue nodes to indicate boys, and purple nodes to indicate girls.

  1. If we have three persons expecting to deliver, what is the probability that at least one of them gives birth to a boy?

    There are two ways to approach this from the diagram. Either find all of the leaves (terminal nodes) in which there is at least one boy, compute the probability of each such outcome by multiplying along the path leading to this leaf (this is Bayes' theorem in disguise), then add up those probabilities (which we can do, because the events are independent).

    Alternatively, we can look for all of the outcomes in which no boys are born, determine the probability of each, and then subtract that probability from $1$ (as all the probabilities must add to $1$). This latter option seems simpler, as there is only one node to consider. Therefore \begin{align} P(\text{at least one boy}) &= 1 - P(\text{no boys}) \\ &= 1 - 0.515^3 \\ &\approx 1 - 0.1366 \\ &= 0.8634, \end{align} which is the answer found in the question.

  2. If we know that at least one will give birth to a boy (suppose we have accurate ultra-sound results), what is the probability all three will have a boy?

    In general, the probability of an outcome $A$, given some event $B$, is $$ P(A \mid B) = \frac{P(A)}{P(B)}.$$ This is, basically, the Law of Total Probability. Take $A$ to be the event "three boys are born" and $B$ to be the event "at least one boy is born". The probability of at least one boy was computed in the first part of the question: $P(B) \approx 0.8634$. Following the same kind of argument as in the first part (i.e. multiply the probabilities along the path from the root node to the leaf labeled "3 boys"), $$ P(A) = P(\text{three boys}) = (0.485)^3 \approx 0.1141. $$

    Then \begin{align} P(\text{three boys} \mid \text{at least one boy}) &= \frac{P(\text{three boys})}{P(\text{at least one boy})} \\ &\approx \frac{0.1141}{0.8634} \\ &\approx 0.1322. \end{align} In other words, if you know that one of the babies is a boy, then the probability that all three are boys is about 13.2%.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.