During the rainy season in a tropical country, the probability of a day being dr

Question

During the rainy season in a tropical country, the probability of a day being dry is 0.10. A dry day is defined as one without significant rainfall. Assuming that the number of days until the next dry day is geometrically distributed, what is the length of consecutive wet days that has a probability of occurence of no more than 0.05? Hint: calculate the probability of one wet day, then, 2, then 3, and so on until you hit a number of consecutive wet days with a probability of occurence of less than 0.05.

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Explanation / Answer

P( day dry)=0.9> P(day is wet)=0.9 P(wet/wet)=0.9*0.9=0.81 wet and before was wet.. P(wet/wet/wet)=0.9*0.9*0.9=0.729 continue same:0.6561, 0.59049, 0.531441, 0.4782969, 0.43046721, 0.387420489... 0.3486784401, 0.31381059609, 0.282429536481, 0.2541865828329, 0.22876792454961, 0.205891132094649, 0.1853020188851841, 0.16677181699666569, 0.150094635296999121, 0.1350851717672992089, 0.12157665459056928801, 0.109418989131512359209, 0.0984770902183611232881, 0.08862938119652501095929, 0.079766443076872509863361, 0.0717897987691852588770249, 0.06461081889226673298932241, 0.058149737003040059690390169, 0.0523347633027360537213511521 NOWWWWWWWWW: 0.04710128697246244834921603689 >>28 days count them again!! PLZ RATE THE MAX it took me much TIME, thx

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