Can you solve this probability question? 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 :)
 A: 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.
