Q. 25

Suppose an integer from 1 through 1000 is chosen at random, fins the probability that the integer is a multiple of 2 or a multiple of 9.

Answer :

Given, Sample space is the set of first 1000 natural numbers.


n(S) = 1000


Let A be the event of choosing the number such that it is multiple of 2


n(A) = [1000/2] = [500] = 500 {where [.] represents Greatest integer function}


P(A) =


Let B be the event of choosing the number such that it is multiple of 9


n(B) = [1000/9] = [111.11] = 111 {where [.] represents Greatest integer function}


P(B) =


We need to find the P(such that number chosen is multiple of 2 or 9)


P(A or B) = P(A B)


Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:


P(E F) = P(E) + P(F) – P(E F)


P(A B) = P(A) + P(B) – P(A B)


We don’t have value of P(A B) which represents event of choosing a number such that number is a multiple of both 2 and 9 or we can say that it is a multiple of 18.


n(A B) = [1000/18] = [55.55] = 55


P(A B) =


P(A B) =


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