Probability of chosen number from 1st 500 divisble by 3or5 , Mathematics

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IN THIS WE HAVE TO ADD THE PROBABILITY of 3 and 5 occuring separtely and subtract prob. of 3 and 5 occuring together therefore p=(166+100-33)/500=233/500=0.466

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3/7/2013 6:18:48 AM s= {1,2,3 ................ 499,500} n(A) = 500/3 = maximum 166 so probability divided by 3 is p(A) = 166/500 n(B) = 500/5 = maximum 100 so probability divided by 5 is p(A) = 10/500 n(A and B) = maximum 33 so probability divided by both 3 and 5 is p(A and B )= 33/500 p(A or B) = p(A)+P(B)-p(A and B) p(A or B) = (166/500)+(100/500)-(33/500) = 233/500

3/7/2013 6:18:48 AM

s= {1,2,3 ................ 499,500}

n(A) = 500/3 = maximum 166 so probability divided by 3 is p(A) = 166/500

n(B) = 500/5 = maximum 100 so probability divided by 5 is p(A) = 10/500

n(A and B) = maximum 33 so probability divided by both 3 and 5 is p(A and B )= 33/500

p(A or B) = p(A)+P(B)-p(A and B)

p(A or B) = (166/500)+(100/500)-(33/500) = 233/500

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