2.1 Practice exercises

2.1.1 Exercises Hard 1 - Hard 3

Solutions by Kent Johnson

Suppose there are two equally common species of panda bear. Species A gives birth to twins 10% of the time, otherwise birthing a single infant. Species B births twins 20% of the time, otherwise birthing singleton infants.

2.1.1.1 2H2: Suppose a female panda of unknown species has just given birth to twins. What is the probability that she is from species A?

The likelihood of birthing twins for each species: \[Pr(twins|A) = 0.1\] \[Pr(twins|B) = 0.2\]

The prior probability that a panda is species A or B: \[Pr(A) = Pr(B) = 0.5\]

The average likelihood of twins is the weighted average of the individual likelihoods. This will be the denominator in Bayes theorem: \[Pr(twins) = Pr(twins|A) * Pr(A) + Pr(twins|B) * Pr(B) \\ = 0.1 * 0.5 + 0.2 * 0.5 \\ = 0.15\]

Bayes theorem gives the posterior probability of A or B given the birth of twins: \[Pr(A|twins) = Pr(twins|A) * Pr(A) / Pr(twins) \\ = (0.1 * 0.5) / 0.15 \\ = 1/3\]

\[Pr(B|twins) = Pr(twins|B) * Pr(B) / Pr(twins) \\ = (0.2 * 0.5) / 0.15 \\ = 2/3\]

2.1.1.2 2H1: What is the probability that her next birth will also be twins?

The posterior probabilities \(Pr(A|twins)\) and \(Pr(B|twins)\) from 2H2 become the prior probabilities for this question. We are asked for the overall probability of twins, this is the weighted average of the probabilities for each species.

These results make sense intuitively–B’s are twice as likely to have twins, so having twins makes them twice as likely to be a B, and moves the probability of twins in the next birth two-thirds of the way from \(Pr(twins|A)\) to \(Pr(twins|B)\).

2.1.1.3 2H3: Suppose the same panda mother has a second birth and that it is a singleton infant. Compute the posterior probability that this panda is species A.

Again, the posterior from problem 2H2 becomes the prior probability of A or B.

Bayes theorem says \[Pr(A|single_twin) = Pr(single|A) * Pr(A|twin) / Pr(single|twin)\]

Substituting: \[Pr(A|single_twin) = (0.9 * (1/3)) / 0.833 = 0.36\]

Check: \(Pr(A|single_twin)\) should be smaller than \(Pr(A|single)\) because the twin birth pushes the probability towards B. \[Pr(A|single) = Pr(single|A) * Pr(A) / Pr(single) \\ = (0.9 * 0.5) / 0.85 \\ = 0.53\]

Notice that observing a twin birth changes the posterior much more than observing a single birth. Twin births are rare and have very different probabilities, so they affect the posterior strongly. Single births are common and have similar probabilities, so they don’t affect the posterior much. ## Meeting Videos

2.1.2 Cohort 1

Meeting chat log

00:54:23    Tony ElHabr:    we need to draw black and white cards to determine who goes
00:54:49    ege can taslicukur: what if i have a card with 2 black faces?
00:54:56    Tony ElHabr:    we end the meeting
00:59:10    ege can taslicukur: 😃
01:05:35    syd235: I think the author just wanted to play a joke on us to make us use pandas in R
01:05:59    Justin Dollman: 🥲