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. [NCERT EXEMPLAR]

Answer 1

Let S be the sample space. Then

Total number of elementary events, n(S) = 1000

Let A be the event that the number selected is a multiple of 2 and B be the event that the number selected is a multiple of 9. Then,

A = {2, 4, 6, ..., 998, 1000}

B = {9, 18, 27, ..., 990, 999}

Now, A ∩ B is the event that the number selected is a multiple of 2 and 9 i.e. 18.

A ∩ B = {18, 36, ..., 990}

We have,

n(A) = $\frac{1000}{2}=500$, n(B) = $\frac{999}{9}=111$ and n(A ∩ B) = $\frac{990}{18}=55$

∴ P(A) = $\frac{500}{1000}$, P(B) = $\frac{111}{1000}$ and P(A ∩ B) = $\frac{55}{1000}$

Now,

P(integer is a multiple of 2 or a multiple of 9)

= P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

= $\frac{500}{1000}+\frac{111}{1000}-\frac{55}{1000}$

= $\frac{556}{1000}$

= 0.556

Hence, the required probability is 0.556.