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

Let S be the sample space. Then Total No. 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 multiple of 9.

Then

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

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

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

$A\cap B$={18, 36,.....990}

We have,

$n\left(A\right)=\frac{1000}{2}=500,n\left(B\right)=\frac{999}{9}=111,$

$n(A\cap B)=\frac{990}{18}=55$

$\therefore P\left(A\right)=\frac{500}{1000},P\left(B\right)=\frac{11}{1000}$

$P(A\cap B)=\frac{55}{1000}$

Now,

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

$=P(A\cup B)$

=$P\left(A\right)+P\left(B\right)-P(A\cap B)$

=$\frac{500}{1000}+\frac{111}{1000}-\frac{55}{1000}=\frac{556}{1000}=0.556$