Suppose there are two species of Tigers, $T_1$ and $T_2$ which are indistinguishable and exist in equal proportions, but differ in how they lay children. Species $T_1$ gives birth to twins 10% of the time and otherwise lays a single cub. Species $T_2$ lays twins 20% of the time, and otherwise only lays a single cub.
There are two tigers, who are unrelated and of unknown species, Tiger X and Tiger Y.
a) Tiger X has twins the first year. Find the conditional probability that the Tiger is species $T_1$.
b) Find the conditional probability that Tiger X will have twins again.
c) Suppose there exists a genetic test which correctly identifies tigers as species $T_1$ 80% of the time and correctly identifies species $T_2$ 60% of the time. This test is administered to Tiger Y, and the results indicate that the tiger is species $T_1$. Find the probability that the first birth from Tiger Y results in twins.
For part a), I was thinking of applying Bayes' theorem to the reduced Universe. Basically speaking $$\frac{0.1}{0.1+0.2}= 0.33$$
However, I'm unsure of how to proceed for parts b and c - I know that I should be using the probability 0.33 as a prior probability, but I'm not sure what the likelihood would be that would allow me to calculate the posterior.
Pretty stumped on c. Would appreciate a detailed explanation. Thanks in advance.
For part (b), you've calculated the probability (based on what you know) that the tiger is type one. You don't need a likelihood, you just need to use the law of total probability with this distribution. So you have $$ P(twins)=P(twins\mid T_1) P(T_1) + P(twins\mid T_2)P(T_2) = \frac{1}{10}\frac{1}{3} + \frac{1}{5}\frac{2}{3} $$ (where I've omitted the other conditioning on the fact they had twins the first time).
Part (c) is Bayes again... this time on a fresh tiger so we can forget everything from (a) and (b). You can use the Bayes rule to get the posterior probability they are type $T_1$ and then use the approach from part (b) to compute the probability the birth is twins. For the first part, $$ P(T_1\mid test) = \frac{P(test\mid T_1)\frac{1}{2}}{P(test\mid T_1)\frac{1}{2} + P(test\mid T_2)\frac{1}{2}} = \frac{\frac{8}{10}\frac{1}{2}}{\frac{8}{10}\frac{1}{2}+\frac{4}{10}\frac{1}{2}}$$ since the probability that it is right given it's a $T_1$ is $8/10$ and the probability the test is wrong given that it's a $T_2$ is $4/10.$