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I want to know if the above question can be solved with Bayes' theorem or if they are independent in the same sense that if it rains today it isn't less likely to rain tomorrow. Is the answer 50%, 33.3...%, or something else?

If you want to lose IQ points feel free to see what everyone answered on 4chan.

I don't think this question is the same as this because 25% doesn't seem to be an option for this question at least in my head.

StubbornAtom
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Jack Pan
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  • related: https://math.stackexchange.com/questions/1266043/boy-and-girl-paradox-is-driving-me-crazy – Henry Mar 04 '20 at 13:58
  • You have to say more! Probably you intend that two hits be independent. In that case, with hit (H) or miss (M) being equally likely with two attempts there are four equally likely out comes: "HH", "HM", "MH", or "MM". Three of those, "HH", "HM", and "MH" have at least one hit. Of those one has two hits so the probability of two hits, given at least one hit, is 1/3. That is the same as Gae S's answer. – user247327 Mar 04 '20 at 14:33

1 Answers1

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The probability is $\frac13$ if the rolls are independent. You are loking at $\frac{P(2\text{ critical hits})}{P(\text{ at least one critical hit})}=\frac{1/4}{3/4}$.

  • From the wording in the image, if that were on a research paper or exam paper, would you assume independence? – Jack Pan Mar 04 '20 at 00:46
  • You are given at least one hit is critical, however, and seek the probability of both being that. – PrincessEev Mar 04 '20 at 00:46
  • @gae-s how does your answer stay consistent to this answer by Kaushik which was voted correct? https://math.stackexchange.com/questions/3446007/what-is-the-probability-of-having-both-girls-given-that-at-least-one-of-them-is edit never mind while the scenario was the same, the actual question was worded differently there – Jack Pan Mar 04 '20 at 00:48
  • @EeveeTrainer True, I corrected. –  Mar 04 '20 at 00:50
  • How about this argument for 50%? @eevie-trainer The "given" critical is either the first or the second. We have two cases then. But they are symmetrical so we only need consider one case, let's say that hit X (1 or 2) is critical. Then hit Y (the other one) still has a 50% chance for a critical so overall the chance is still 50% no? – Jack Pan Mar 04 '20 at 00:52
  • And what about the argument for 25% that since "at least 1" is contained in "both are critical", the probability is determined at the beginning anyway? – Jack Pan Mar 04 '20 at 01:04
  • @MaxLi Should there actually be two critical hits, which would be the "given" critical hit and which the "other one"? – Graham Kemp Mar 04 '20 at 02:18
  • @MaxLi "At least one" is not contained in "both are critical". It is the converse. "Both are critical" is a subset of "at least one is critical". Hence the answer by Gae. S. holds. – Graham Kemp Mar 04 '20 at 02:25
  • @MaxLi In the argument for $0.5$ you are saying that $\frac{P(A_1\cap A_2)}{P(A_1)}=0.5$, that $\frac{P(A_1\cap A_2)}{P(A_2)}=0.5$, and then how do you get to $\frac{P(A_1\cap A_2)}{P(A_1\cup A_2)}=\text{stuff}$? –  Mar 04 '20 at 10:20