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From experience, the manager of Kramer's Book Mart knows that 30% of the people who are browsing in the store will make a purch...Asked by titi
                From experience, the manager of Kramer's Book Mart knows that 50% of the people who are browsing in the store will make a purchase. What is the probability that among ten people who are browsing in the store, at least two will make a purchase? (Round your answer to four decimal places.)
            
            
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                    Answered by
            Gursafal
            
    the probability that n person buy the book is (0.5)^n
so the probability that at least two people buy a book is (0.5)^2 + (0.5)^3 ....... (0.5)^10
which is equal to 0.498046875
    
so the probability that at least two people buy a book is (0.5)^2 + (0.5)^3 ....... (0.5)^10
which is equal to 0.498046875
                    Answered by
            MathMate
            
    Again here the keyword is "among" 10 people.
Check that the conditions of binomial distribution are satisfied (ref. the airplane problem).
Then proceed with the calculation of "at least two" out of 10, namely "not zero", "not one" out of 10.
Instead of calculating B(2,10,0.5)+B(3,10,0.5)+...+B(10,10,0.5), we calculate the complement, i.e.
P(x>=2)
=1-P(x<)
=1-(P(x=0)+P(x=1))
=1-(B(0,10,0.5)+B(1,10,0.5))
See the airplane problem for the formula for B(x,n,p):
https://www.jiskha.com/display.cgi?id=1499053975
    
Check that the conditions of binomial distribution are satisfied (ref. the airplane problem).
Then proceed with the calculation of "at least two" out of 10, namely "not zero", "not one" out of 10.
Instead of calculating B(2,10,0.5)+B(3,10,0.5)+...+B(10,10,0.5), we calculate the complement, i.e.
P(x>=2)
=1-P(x<)
=1-(P(x=0)+P(x=1))
=1-(B(0,10,0.5)+B(1,10,0.5))
See the airplane problem for the formula for B(x,n,p):
https://www.jiskha.com/display.cgi?id=1499053975
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