Question

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

Answer 1

We have multiples of 2, from 1 to 1000 are 2, 4, 6, 8, ..., 1000

Let n be the number of terms

a = 2, d = 2, a_n = 1000

a_n = a + (n – 1)d

1000 = 2 + (n – 1)2

⇒ 1000 = 2n

⇒ n = 500

Now multiple of 9 from 1 to 1000 are 9, 18, 27, ..., 999

Here a = 9, d = 9 and a_m = 999   .....[m is the number of terms]

a_m = a + (m – 1)d

⇒ 999 = 9 + (m – 1)9

⇒ 999 = 9m

⇒ m = 111

Now number multiples of 2 and 9 are 18, 36, 54, ..., 990

Here a = 18, a_p = 990, d = 18   .....[p is the number of terms]

∴ a_p = a + (p – 1)d

990 = 18 + (p – 1)18

⇒ 990 = 18 + 18p – 18

⇒ 18p = 990

⇒ p = 55

∴ Number of multiples of 2 or 9 = 500 + 111 – 55

= 556

∴ Required probability = `(P(E))/(n(E))`

= `556/1000`

= 0.556

Hence, the required probability = 0.556