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]

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

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