First off let me say I love math, I love probabilities, quizzes, IQ tests, ... This doesn't mean I'm great at it :)

To help improve this, I usually download math related apps on my mobile phone to keep me entertained on the train. One of these apps contains "probability puzzles". One of the easy puzzles states the following:

Suppose babies are boys with probability 0.51, girls with probability 0.49, and births are independent.

If you have two children, what's the probability they're both girls?

This seemed straight forward to me.

  • 0 girls = 0.255
  • 1 girl = 0.49
  • 2 girls = 0.245

But the puzzle indicates this as wrong. Am I missing something? The only thing I can't quite set into context is the following sentence from the question:

and births are independent

What does this actually mean? (not a native English speaker).

Thank you

  • 1
    $\begingroup$ Births being independent refers to Independence (probability theory). The short explanation is that $A$ and $B$ are independent events iff $P(A\cap B)=P(A)\cdot P(B)$ iff $P(A\mid B)=P(A)$, i.e. the probability of $A$ occurring does not at all depend on whether or not $B$ has occurred, i.e. whether we have knowledge that $B$ as having occurred or not we cannot know anything more about the probability of whether or not $A$ has occurred. $\endgroup$ – JMoravitz Nov 15 '16 at 8:28
  • 1
    $\begingroup$ $0$ girls would be $0.51^2$, $1$ girl would be $0.51\cdot 0.49\cdot 2$ and $2$ girls would be $0.49^2$. See binomial distribution. $\endgroup$ – JMoravitz Nov 15 '16 at 8:29
  • $\begingroup$ Oh waw, was I really that far off?! Now I feel stupid :) $\endgroup$ – Goowik Nov 15 '16 at 8:42

" births are independent" intuitively mean that the gender of the previous child does not affect the gender of the next child.

In mathematical language, if $A$ and $B$ are independent events, then $P(A\cap B)=P(A)\times P(B)$.

So in this case the probability of having two girls should be (0.49)(0.49)=0.2401.

  • 2
    $\begingroup$ okay thanks for clearing up (to all of you). Not sure why I was thinking of dividing it. But now the independence seems clear. It's like drawing an ace from 2 different decks of cards (instead of one). $\endgroup$ – Goowik Nov 15 '16 at 8:44

You have $4$ disjoint events:

  • BB, for which the probability is $0.51\cdot0.51=0.2601$
  • BG, for which the probability is $0.51\cdot0.49=0.2499$
  • GB, for which the probability is $0.49\cdot0.51=0.2499$
  • GG, for which the probability is $0.49\cdot0.49=0.2401$
  • $\begingroup$ I see thanks. I was confused at first. $\endgroup$ – Goowik Nov 15 '16 at 8:58

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