https://notexponential.com/qa/artificial-intelligence/probability Powered by Question2Answer https://notexponential.com/763/disco-fever-and-two-kinds-of-tests?show=968#a968 P(RAT)=2/3 &nbsp;&nbsp;&nbsp;P(FST)=1/3 &nbsp;P(disease)=1/4<br /> <br /> RAT =&gt; P(+ve|disease) = 3/4 &nbsp;&nbsp;&nbsp;P(+ve|Nodisease) = 1/4<br /> <br /> FST =&gt; P(+ve|disease) = 1 &nbsp;&nbsp;&nbsp;P(+ve|Nodisease) = 1/3<br /> <br /> P(disease|+ve) = P(+ve|disease)*P(disease) / [P(+ve|diasease) + P(+ve|Nodisease)] <br /> <br /> = [[P(RAT)*P(+ve|disease)+P(FST)*P(+ve|disease)]*P(disease)] / [[P(RAT)*P(+ve|disease) +P(FST)*P(+ve|disease)]*P(disease)+[P(RAT)*P(+ve|Nodisease)+P(FST)*P(+ve|Nodisease)]*P(Nodisease)]<br /> <br /> = [[2/3*3/4+ 1/3*1]*1/4] / [[2/3*3/4 + 1/3*1]*1/4 + [2/3*1/4 + 1/3*1/3]*3/4]<br /> <br /> = [1/8 + 1/12] / [1/8 + 1/12 + 1/8 + 1/12]<br /> <br /> = [1/8 + 1/12] / 2*[1/8 + 1/12]<br /> <br /> =1/2<br /> <br /> Therefore, probability the person has disco fever given the result is positive is 0.5 Probability https://notexponential.com/763/disco-fever-and-two-kinds-of-tests?show=968#a968 Sat, 04 May 2024 22:13:48 +0000 https://notexponential.com/799/coin-real-or-biased?show=801#a801 Let&#039;s define the following events:<br /> <br /> A: Dealer is using the fair coin.<br /> <br /> B: Dealer is using the biased coin.<br /> <br /> O: We observe the sequence H, H, H.<br /> <br /> We want to calculate the probability of event B given O, i.e., P(B|O).<br /> <br /> By Bayes&#039; theorem, we have: P(B|O) = P(O|B) * P(B) / P(O)<br /> <br /> P(O|B) = 1<br /> <br /> P(B) = 0.5<br /> <br /> P(O) = P(O|A)*P(A) + P(O|B)*P(B)<br /> <br /> We need to calculate P(O|A) and P(O|B):<br /> <br /> P(O|A) = P(H, H, H|A) = P(H|A)^3 = 0.5^3 = 0.125<br /> <br /> P(O|B) = P(H, H, H|B) = 1<br /> <br /> P(O) = P(O|A)*P(A) + P(O|B)*P(B) = 0.125 * 0.5 + 1 * 0.5 = 0.5625<br /> <br /> Now we can use these values for the Bayes theorem:<br /> <br /> P(B|O) = P(O|B) * P(B) / P(O) = 1 * 0.5 / 0.5625 = 0.8889 Probability https://notexponential.com/799/coin-real-or-biased?show=801#a801 Tue, 25 Apr 2023 18:42:20 +0000 https://notexponential.com/753/probability-rain-desert-reliable-weatherperson-predicts?show=791#a791 Let&#039;s say that R is the chance of it raining on a given day and FR is the chance of the weather forecaster calling for rain on a given day.<br /> <br /> From Bayes&#039; Theorem, P(R|FR) = [P(FR|R)*P(R)]/P(FR). In other words, the probability of it raining given a forecast for rain is [ (probability of the forecast for rain on days it actually rains) * (probability of it raining on a given day) ] / (the probability of the forecaster saying it will rain on a given day).<br /> <br /> P(FR|R) = 0.9 (since it rains 90% of the time a forecaster predicts rain)<br /> <br /> P(R) = 5/365 (days of the year it should rain)<br /> <br /> We need to consider two cases for P(FR) and sum them together:<br /> Case 1: The forecaster calls for rain on days it will rain<br /> Case 2: The forecaster calls for rain on days it will not rain<br /> <br /> For Case #1, the forecaster is right about 90% of the days it will rain. Thus, <br /> P( Case 1 ) = 0.9 * 5/365.<br /> <br /> For Case #2, the forecaster calls for rain on 10% of the days it won&#039;t actually rain. Thus,<br /> P( Case 2 ) = 0.1 * 360/365<br /> <br /> P(FR) = P( Case 1 ) + P( Case 2 )<br /> <br /> Thus,<br /> <br /> P(R|FR) = (0.9 * (5/365)) / ( (0.9 * (5/365)) + (0.1 * (360/365)) ) = 0.111...<br /> <br /> Thus, if the forecaster predicts rain, it will rain ~11.1% of the time. Probability https://notexponential.com/753/probability-rain-desert-reliable-weatherperson-predicts?show=791#a791 Sat, 04 Mar 2023 21:17:45 +0000