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The Koozebanian Fazoob has three genders, ma, na and muna, each with equal probability of being born. Fazoobs are oviparous and large clutches of eggs are incubated in hatching cells. During one particular day, in a certain cell, one hundred eggs hatch. There are twenty more ma hatchlings than na hatchlings and the total number of ma and na hatchlings is four times larger than the number of muna hatchlings. One more egg hatches overnight and in the morning a brood mother selects one of the one hundred and one hatchlings at random for cleaning; it happens to be a ma. What is the probability that the egg that hatched overnight contained a ma? Please help me with this question :)

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    Yes We Can!!!!! – 2017-02-15
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    @barakmanos I've thought about the same answer :D HappyManager - please, show us what have you done so far. – 2017-02-15
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    I thought it would be 1/3, since each gender has an equal probability of being born but it turns out it isn't! – 2017-02-15
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    I know there are 50 ma, 30 na and 20 muna – 2017-02-15
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    In the sample size of a 100 they are divided into 50,30,20 – 2017-02-15
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    Yeah, I read the question more carefully the second time. Anyways, have you heard of Bayes' law? Plug in the numbers there, and the problem is solved. – 2017-02-15
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    Also, I'll tell you why the answer isn't a third with an example: Imagine the hundred eggs were a 50-50 split between na's and muna's (and only those). In that case, the probability that the one additional egg had a ma in it is clearly $1$. The same thing happens in your problem, only less extreme. – 2017-02-15
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    Bayes law seems really complicated- I do not see how it correlates with the question? – 2017-02-15
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    Someone please help – 2017-02-15
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    [Contest question](http://www.maths.manchester.ac.uk/mathsbombe/problem.php?index=5) – 2017-02-15

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Looking at the two events "A ma hatched" (hereby called $h$) and "A ma was picked" ($p$), the formula for conditional probability yields $$ P(h\mid p) = \frac{P(h\cap p)}{P(p)} $$ So, we need to calculate those two probabilities. The numerator $P(h\cap p)$ first: By the product rule for consecutive events, we get $$ P(h\cap p) = P(h)\cdot P(p\mid h) = \frac13 \cdot \frac{51}{101} = \frac{17}{101} $$ The denominator next. By the law of total probability, we get $$ P(p) = P(p\mid h)\cdot P(h) + P(p\mid \lnot h)\cdot P(\lnot h)\\ = \frac{51}{101}\cdot\frac{1}{3} + \frac{50}{101}\cdot \frac23\\ = \frac{151}{303} $$ Now we're ready to calculate the conditional probability: $$ P(h\mid p) = \frac{P(h\cap p)}{P(p)} = \frac{17/101}{151/303} = \boxed{\!\frac{51}{151}} $$ You will note that this is slightly higher than $\frac13 = \frac{51}{153}$. Basically, that's accounting for the possibility that the brood mother picked the last egg. She probably didn't, which is why it's still close to $\frac13$, but she might have, which is why the final probability is slightly above.

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    I started doing Bayes' theorem, but then I realised doing conditional probability from the ground was just as easy. – 2017-02-15