Question

Suppose the number of cars in a household has a binomial distribution with parameters n = 12, and p = 10 %. Find the probability of a household having: (a) 1 or 5 cars (b) 3 or fewer cars (c) 9 or more cars (d) fewer than 5 cars (e) more than 3 cars

Answer 1

			Probability
(a)	P(1 or 5 cars)	P(1)+P(5)	0.38036082676
(b)	P(3 or fewer cars)	P(0)+P(1)+P(2)+P(3)	0.97436252984
(c)	P(9 or more cars)	P(9)+P(10)+P(11)+P(12)	0.00000016584
(d)	P(fewer than 5 cars)	P(0)+P(1)+P(2)+P(3)+P(4)	0.99567065673
(e)	P(more than 3 cars)	P(4)+P(5)…......+P(11)+P(12)	0.02563747017

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