Question

What is the probability that a number selected at random from the first 50 positive integers is exactly divisible by 3 or 4?

Answer 1

Sample space (N(S)) = 50

Let A be the event of getting a number divisible by 3

A = {3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48}

N(A) = 16

Therefore, the probability of a number divisible by 3 is

`P(A)=(N(A))/(N(S))=(16)/(50)`

Let B be the event of getting a number divisible by 4

B = { 4,8,12,16,20,24,28,32,36,40,44,48}

N(B) = 12

Therefore, the probability of a number divisible by 4 is

`P\left(B\right)=(N(B))/(N(S))=(12)/(50)`

Also, the intersection between A and B is

A n B = { 12,24,36,48}

Therefore,

`P\left(A\cap B\right)=(N(A\cap B))/(N(S))=(4)/(50)`

Hence, the probability that a number selected at random from the first 50 positive integers is exactly divisible by 3 or 4 will be

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

Therefore,

`\begin{gathered} P(A\cup B)=(16)/(50)+(12)/(50)-(4)/(50)=(24)/(50)=(12)/(25) \\ \therefore P(A\cup B)=(12)/(25) \end{gathered}`

Hence, the answer is

`(12)/(25)`