New answers tagged

4

Same answer, different proof. Proof by induction on $n$. Then assume it is true for all $k: 1 \leq k \leq n-1$. Obviously: Clearly, QED


-1

Given the wording of "whose turn it is", we are left to assume that taking out the trash will be a recurring event. In this way, we can guarantee fairness regardless of $n$ and $p_i$ by simply recording who did it last and alternating each time. The coin flipping only needs to occur for the very first time, but on a sufficiently large timeline, one sample ...


5

Since we are talking about multiplication (products grow really fast), and our data gathering is relatively cheap, it stands to reason that the best strategy is to I had this idea yesterday, but it took me this long to get through the final annoying special case. However, I'm pretty convinced now that This is going to be a longish read because of the ...


9

A set of coins is fair in the relevant sense if and only if Proof (slightly highbrow, sorry): Alternative kinda-equivalent proof (simpler ideas but needs you to know a theorem): (Neil W suggested, in comments, taking that second approach. I'd avoided it because the other way seemed quicker and more first-principles-y, but the second way may well be easier ...


0

Because of the codominancy of the mother's parents, the mother is BB pure blood. We only need to calculate with the father this time. We don't know anything about the fathers parents so the father can be any fenotype with equal probability, from this point of view. But the starting point in the question is that we already know that I am B fenotype, so ...


3

As JNF says, your mother must have AB parents and be BB herself. Applying the final point directly to your father, before taking into account information from your own blood type your priors for his genotype are OO: $0.25$, AB: $0.25$, AA: $0.125$, AO: $0.125$, BB: $0.125$, BO: $0.125$. Thus the probabilities of a given allele from your father (again ...


1

Mom's parents have to be Meaning mother is So options for father are So it seems


0

The answer is not 60% it is The parents have a 25% chance being BBBB, 50% chance being BBBO and 25% chance being BOBO. This should not change when eliminating OO. In the case of BBBB, the child is BB for sure, which is 100% chance. In the case of BBBO, where one parent is BB and the other is BO, the child is 50% BB depending on whether the BO parent gives ...


3

My parents' blood types can be one of four scenarios, which are equally-likely since they are not conditioned on my own blood type (clarified in comments): How likely is it for me to have inherited the different possibilities, under each of these parental scenarios? Multiplying by 4 to turn these into integers instead of percentages, we get the following ...


1

According to: it is: Q2:


1

I think that where you're getting confused is in the framing of the question. The question is asking for the fewest number of candies we need to take in order to guarantee that 7 candies share a colour. The scenario you gave results in 7 candies sharing a colour after picking 13 candies out. However, if I can find a scenario where after picking 13 or more ...


Top 50 recent answers are included