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 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.