First, I know this riddle has been asked (many times) before. The question I want answering is why is a tree diagram not a correct method for determining the probability in this case.

There are two children, equally likely to be Boy or Girl. If we know one (or more) is a Boy, what is the probability that there is a Girl in the pair?

The sample space looks like this:

GG - not possible

Therefore the probability is 2/3, but if I draw a tree diagram:

tree diagram

The probability of a girl in the pair seems to be $$\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$$

Why does a tree diagram fail?


Your probiabilities are not correct. Note that if we call the two children $X_1$ and $X_2$, then \begin{align*} \def\P{\mathbf P}\P[X_1 = B \mid X_1 = B \vee X_2 = B] &= \frac{\P[X_1 = B]}{\P[X_1 = B \vee X_2 = B]}\\ &= \frac{\frac 12}{\frac 34}\\ &= \frac 23 \end{align*} So, the probiabilities at the first node should be $\frac 23$ and $\frac 13$, respectively, giving $$ \frac 23 \cdot \frac 12 + \frac 13 = \frac 13 + \frac 13 = \frac 23 $$ in total.

  • $\begingroup$ That's a neat equation, does it have a name? $\endgroup$ – Jamie Twells Mar 15 '16 at 12:02

In your tree diagram, the $G-B$ link should also have a probability of $\dfrac12$,
and the $G-G$ link should be crossed out.

So $Pr = \dfrac{1/4 + 1/4}{3/4} = \dfrac23$


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.