Question

One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

Answer 1

Let S be the sample space.
Then n(S) = 100
∴ Total number of elementary events = 100
Let A be the event where the number selected is divisible by 4 and B be the event where the number selected is divisible by 6.
Then A = {4, 8, 12, 16, ...100 }
B = { 6, 12, 18, 24, ...96},
and (A ∩ B) = {12, 24, ...96}
Now, we have:
$n\left(A\right)=\frac{100}{4}=25$

$n\left(B\right)=\frac{96}{6}=16$

$n\left(A\cap B\right)=\frac{96}{12}=8$ [∵ LCM of 4 and 6 is 12]

$\therefore P\left(A\right)=\frac{25}{100},P\left(B\right)=\frac{16}{100}\mathrm{and}P\left(A\cap B\right)=\frac{8}{100}$

Now, required probability = P(a number is divisible by 4 or 6)
= P (A ∪ B)
= P(A) + P(B) $-$P(A ∩ B)
= $\frac{25}{100}+\frac{16}{100}-\frac{8}{100}=\frac{25+16-8}{100}=\frac{33}{100}$