A family has 3 children, creatively named A, B, and C.

(a) Discuss intuitively (but clearly) whether the event "A is older than B" is independent of the event "A is older than C."

(b) Find the probability that A is older than B, given that A is older than C.

(a) Discuss intuitively (but clearly) whether the event "A is older than B" is independent of the event "A is older than C."

(b) Find the probability that A is older than B, given that A is older than C.

Solution: (a) They are not independent: knowing that A is older than B makes it more likely that A is older than C, as the if A is older than B, then the only way that A can be younger than C is if the birth order is CAB, whereas the birth orders ABC and ACB are both compatible with A being older than B. Consult iTunes for full solution. (b) By conditioning and Bayes' Rule, 2/3. Note that unconditionally, any of the 3 children is equally likely to be the eldest.

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