Question 909197: The probability of rain on a given day in September is 0.25.
a) Assuming independent trials, what is the probability that there will be no rain over the three days of the Labour Day Weekend?
b) Assuming independent trials, what is the probability that it will ran at least once in a given week?
c) Assuming independent trials, what is the probability that it will rain twice in one week?
d) What is the mean (i.e, expected) number of rainy days in September?
e) Is the assumption of independent trials in this example realistic?
I am still new at these concepts and am having trouble solving. Any help would be greatly appreciated.
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! Let X~Binomial(n=3,p=.25)
a) P[X=0] = (3 choose 0)(.25)^0*(.75)^3 = .4219
Let X~Binomial(n=7, p=.25)
b) P[x>=1] = 1-P[X=0] = 1-(.75)^7 = .8665
c) P[X=2] = (7 choose 2)*(.25)^2*(.75)^5 = .3115
Let X~Binomial(n=30, p =.25) [30 days in Sept]
d) Mean of a binomial = n*p.
30 * .25 = 7.5
e) I would say that assumption of independent trials is realistic. Just cause it rains today doesn't significantly change the probability of it raining the next day as well.
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