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  • 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.

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.

Answers (1)

Given: we have integers 1, 2,..., 1000

No. of outcomes, n(S) = 1000

No. of the integers that are multiples of 2

2, 4, 6, 8,... 1000.

Let us consider ‘p’ as the number of integers.

Now, ap = a + (p - 1)d

On substituting the values, we get,

2 + (p-1)2 = 1000

2 + 2p – 2 = 1000

Thus, p = 1000/ 2

Thus, p = 500

Thus, the number of integers that are multiples of 2 = 500

No. of the integers that are multiples of 9

9, 18, 27, 35,..., 999.

Let us consider ‘n’ as the number of integers.

Now, an = a + (n-1)d.

On substituting the values, we get,

9 + (n - 1)9 = 999

9 + 9n - 9 = 999

Thus, n = 999/9.

Thus, n = 111

Thus, the number of integers that are multiples of 9 = 111

Now, let m be the number of multiples common for both 2 & 9, viz. 18, 36,..., 990.

Thus, the nth term will be 990

Now, am = a + (m - 1)d

We know that. a = 2 & d = 9

Substituting the respective values, we get,

18 + (m - 1)18 = 990

18 + 18m – 18 = 990

Thus, m = 990/18

Thus, m = 55

Now, the number of multiples of 2 or 9 will be:

No. of multiples of 2 + no. of multiples of 9 – No. of multiples of both 2 & 9

= 500 + 111 – 55

= 556

= n(E)

Required probability = number of favourable outcomes/total number of outcomes

= n(E) / n(S)    

= 556 / 1000

= 0.556

Posted by

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