We toss a biased (an unfair) coin 10 times. The coin has a probability of tossing heads p. Calculate the probability that there are 5 heads in the first 8 tosses and 3 heads in the last 5 tosses (both of them must happen), in terms of p. [Solved by HardCarry]
Suppose there is a crazy professor who grades some submissions with marks {1, 2, 3, 4, 5, 6} and he does it totally randomly. How many times is the mean value of submissions, in which you will have every mark at least once? [Solved by HardCarry]
Suppose there are 2n persons who form n couples. Suppose that after many years, the probability of a person being alive is p, and it is equally likely for all persons. On condition that after many years, m people are alive, find the mean value of couples who have both persons alive. (on terms of m and n)
deletedabout 7 years
In a class I taught at Berkeley, I did an experiment where I wrote a simple little program that would let people type either “f” or “d” and would predict which key they were going to push next. It’s actually very easy to write a program that will make the right prediction about 70% of the time. Most people don’t really know how to type randomly. They’ll have too many alternations and so on. There will be all sorts of patterns, so you just have to build some sort of probabilistic model. Even a very crude one will do well. I couldn’t even beat my own program, knowing exactly how it worked. I challenged people to try this and the program was getting between 70% and 80% prediction rates. Then, we found one student that the program predicted exactly 50% of the time. We asked him what his secret was and he responded that he “just used his free will.”
I calculated the probability that it rains TODAY. Are you asking for the probability that it rains EVERYDAY?
If the latter, then you are correct. It's independent and doesn't matter. We know nothing and we can't say anything about the probability that it rains. If the former, then I should be right.
Right, you don't know what the next flip is because you don't know what the person with 100% accuracy is going to guess next. Those 2 events are independent. But you DO know what the current flip is going to be because you DO know what the person with 100% accuracy is guessing currently. These 2 events are not independent and your weatherman problem is the latter, not the former case.
We don't know the probability of rain, but we can get a probability in terms of what the probability of rain usually is (p). I don't try to say that I know the probability of rain, I put my answer in terms of p so that if we did know it, we could get the probability.
We don't need to know the probability that the predictions are actually correct, we only need to know the probability that the predictions are "It Rains" which is different from the predictions being correct.
deletedabout 7 years
guys just wait until tomorrow and actually see if it rains
deletedabout 7 years
I'm on my phone and can't fix a few errors I see in there, oh well.
deletedabout 7 years
Your problem is that you're using misconstruing when Bayes should be used. You recognize that the problem can have all of its information fit into that model, but fail to see that using Bayes would actually net you some weird answer. My comparison of the coin flip was to say that: no matter what, the chances of heads are 50% - so even if someone has a 100% prediction rate does not mean that the next flip will be heads.
You don't know the probability of rain. You don't know the probability that the predictions are actually correct (see wording) You don't know the probability of anything given the other occurs
It's a honey pot of misinformation that is worded like a common textbook problem
Yeah, exactly. What i'm saying is that you can't solve the problem without knowing what the actual chances of rain are. The problem is unsolvable without this information. You can only get an answer in terms of p, the actual chances of rain.
The probability of it raining is 0.28p/(0.1+0.18) with p being the actual chances of rain. You can't go further than that, but I think my usage of Bayes is fine. What I'm trying to say is that you can use Bayes to get that formula and that I didn't make any incorrect assumptions about what the weatherman were basing their predictions on . You just can't go further than that.
If the chance of raining is 50% usually and a weatherman with 100% accuracy said that it is raining,
The probability is no longer just 50%, it's 100% because you know that the chances must be higher because of the additional information you are given about the weatherman's prediction.
In your coin example, the additional information doesn't have any effect on the original probability but in this case and in your original weatherman problem, it does. The 2 events are not independent, so you can use Bayes to make a conclusion.
I might be wrong, but I don't think that's the same situation. The additional information you give in the weatherman problem is sufficient to reduce your probability space to only Cases 1 and 5. In your coin situation, whatever people guessed is INDEPENDENT of the result of the next coin flip, so it makes no difference. The probability of it landing tails will still be 50%.
In the weathermen problem, you're asking the probability of it raining originally which is a conditional probability. Whatever the weatherman guessed DOES have an effect on the probability of it raining. P(A|B) is not equivalent to P(A) as it is in the coin example you just gave.
deletedabout 7 years
Just because there's x number of outcomes doesn't mean it's the correct way to solve it, look at it this way:
If person A has a 90% prediction rate on guessing what side a coin flip will land on, and person B has a 20% prediction rate and they both guess heads
What are the chances that there next coin flip you make will land on tails?
See the difference?
deletedabout 7 years
I can't even do simple math reflex let alone this.
I don't think I'm assuming that the observed predictions of the weathermen were based on datasets of rain not given. It's just 8 possibilities:
1. It rains, Weatherman A predicts rain (he is right), Weatherman B predicts rain (he is right) 2. It rains, Weatherman A predicts rain (he is right), Weatherman B does not predict rain (he is wrong) 3. It rains, Weatherman A does not predict rain (he is wrong), Weatherman B predicts rain (he is right) 4. It rains, Weatherman A does not predict rain (he is wrong), Weatherman B does not predict rain (he is wrong)
5. It does not rain, Weatherman A predicts rain (he is wrong), Weatherman B predicts rain (he is wrong) 6. It does not rain, Weatherman A predicts rain (he is wrong), Weatherman B does not predict rain (he is right) 7. It does not rain, Weatheman A does not predict rain (he is right), Weatherman B predicts rain (he is wrong). 8. It does not rain, Weatherman A does not predict rain (he is right), Weatherman B does not predict rain (he is right).
The probability tree I drew finds the probability of all 8 events happening. Given the information we are given, we can restrict our probability to only Cases 1 and 5.
The probability we are looking for is Case 1 / (Case 1 + Case 5)
deletedabout 7 years
Technically correct but still wrong.
You can't solve that problem, or prove that's it's not able to be solved with Bayes as you have no event information, only hypothesis. You made the mistake of assuming that the observed predictions of the weather men were based on datasets of rain not given here, which could be wrong.
Replace weathermen with psychics and you would probably sniff the answer to this question out sooner. This type of logic is used in sniffing out potential failed investments: just because one has been met with previous success before does not mean that the trend will arbitrarily continue.
I don't think I understand what Question 3 is saying. Could you elaborate a bit? I'm not quite sure how you can put p in terms of m and n in the current wording of the question.
