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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