# Suppose an integer from 1 through 1000 is chosen at random,

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.

Solution:

We have integers 1, 2, 3 … 1000

∴ Total number of outcomes, n(S) = 1000

Number of integers which are multiple of 2 are 2, 4, 6, 8, 10, … 1000

Let p be the number of terms

We know that, ap = a + (p – 1) d

Here, a = 2, d = 2 and ap = 1000

Putting the value, we get

2 + (p – 1)2 = 1000

⇒ 2 + 2p – 2 = 1000

p = 1000/2

⇒ p = 500

Total number of integers which are multiple of 2 = 500

Let the number of integers which are multiple of 9 be n.

Number which are multiples of 9 are 9, 18, 27, …999

∴ nth term = 999

We know that, an = a + (n – 1) d

Here, a = 9, d = 9 and an = 999

Putting the value, we get

9 + (n – 1)9 = 999

⇒ 9 + 9n – 9 = 999

n = 999/9

⇒ n = 111

So, the number of multiples of 9 from 1 to 1000 is 111.

The multiple of 2 and 9 both are 18, 36, … 990.

Let m be the number of terms in above series.

∴ mth term = 990

We know that, am = a + (m – 1) d

Here, a = 9 and d = 9

Putting the value, we get

18 + (m – 1)18 = 990

⇒ 18 + 18m – 18 = 990

m = 990/18

⇒ m = 55

Number of multiples of 2 or 9

= No. of multiples of 2 + no. of multiples of 9 – No. of multiples of 2 and 9 both

= 500 + 111 – 55

= 556 = n (E)

Required Probability $=\frac{\text { Number of favourable outcome }}{\text { Total number of outcomes }}$

$=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$

$=\frac{556}{1000}$

$=0.556$