Question

An integer is chosen at random from first 200 positive integers. Find the probability that the integer is divisible by 6 or 8.

Answer 1

Let S be the sample space. Then n(S) = 200
∴ Total number of elementary events = 200
Let A be the event in which the number selected is divisible by 6 and B be the event in which the number selected is divisible by 8.
Then A = {6, 12, 18, 24, ...198 },
B = { 8, 16, 24, 32, ...200}
and (A ∩ B) = {24, 48, 72, ...192}

Now, we have:
$n\left(A\right)=\frac{198}{6}=33$
$n\left(B\right)=\frac{200}{8}=25$
$n\left(A\cap B\right)=\frac{192}{24}=8$ [∵ LCM of 6 and 8 is 24]

$\therefore P\left(A\right)=\frac{33}{200},P\left(B\right)=\frac{25}{200}\mathrm{and}P\left(A\cap B\right)=\frac{8}{200}$

Now, required probability = P(a number is divisible by 6 or 8)
= P (A ∪ B)
= P(A) + P(B) $-$ P(A ∩ B)
= $\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{33+25-8}{200}=\frac{50}{200}=\frac{1}{4}$