Question

One number is to be chosen from numbers $1$ to $100$, the probability that it is divisible by $4$ or $6$ is:
A.$\dfrac{{33}}{{100}}$
B.$\dfrac{7}{{100}}$
C.$\dfrac{4}{{100}}$
D.$\dfrac{{43}}{{100}}$

Answer 1

Hint: Here, we will first find the numbers that are divisible by 4 , 6 and both 4 and 6. Then by using the probability formula, we will find the probability that the number is divisible by 4, 6 and both. We will use the OR rule to find the probability that is divisible by either 4 or 6. Probability is defined as the certainty of occurrence of an event and it always lies between 0 to 1.

Formula Used:
 We will use the following formulas:
1.The probability is given by the formula $P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$ where $n\left( A \right)$ is the number of favorable outcomes and $n\left( S \right)$ is the total number of outcomes.
2.OR rule: $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

Complete step-by-step answer:
We are given the numbers from 1 to 100.
So, the Sample Space, $n\left( S \right) = 100$
Let A be the event of choosing a number from 1 to 100 that are divisible by 4
Numbers from 1 to 100 divisible by 4
 $A = \left\{ {4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100} \right\}$
Number of Numbers from 1 to 100 divisible by 4 is $n\left( A \right) = 25$.
Substituting $n\left( S \right) = 100$ and $n\left( A \right) = 25$ in the formula of probability,$P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$, we get
Probability of choosing a number from 1 to 100 divisible by 4, $P\left( A \right) = \dfrac{{25}}{{100}}$
Let B be the event of choosing a number from 1 to 100 that are divisible by 6
Numbers from 1 to 100 divisible by 6
$B = \left\{ {6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96} \right\}$
 Number of Numbers from 1 to 100 divisible by 6 is $n\left( B \right) = 16$ .
Substituting $n\left( S \right) = 100$ and $n\left( B \right) = 16$ in the formula of probability $P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}}$, we get
So, probability of choosing a number from 1 to 100 divisible by 6 is
$P\left( B \right) = \dfrac{{16}}{{100}}$
Now, we will find the probability that the chosen number is divisible by both 4 or 6
 Numbers from 1 to 100 that are divisible by 4 and 6:
$A \cap B = \left\{ {12,24,36,48,60,72,84,96} \right\}$
Number of Numbers from 1 to 100 that are divisible by 4 or 6:
$n\left( {A \cap B} \right) = 8$
Now substituting $n\left( S \right) = 100$ and $n\left( {A \cap B} \right) = 8$ in the formula of probability $P\left( {A \cap B} \right) = \dfrac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}}$, we get
So, the probability of choosing a number from 1 to 100 divisible by 6:
$P\left( {A \cap B} \right) = \dfrac{8}{{100}}$
Now, by using the OR rule in probability $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ and substituting the values, we get
$P\left( {A \cup B} \right) = \dfrac{{25}}{{100}} + \dfrac{{16}}{{100}} - \dfrac{8}{{100}}$
Adding and subtracting the terms, we get
$ \Rightarrow P\left( {A \cup B} \right) = \dfrac{{33}}{{100}}$
Therefore, the probability that it is divisible by 4 or 6 is $\dfrac{{33}}{{100}}$ .
Thus Option (A) is the correct answer.

Note: We know that the OR rule in probability states that the outcome has to satisfy one condition or other condition or both the conditions at the same time. If the event is a mutually exclusive event, then the probabilities of one condition and the other condition has to be added, so the event is also called a disjoint event. AND rule in probability states that that the outcome has to satisfy both the conditions at the same time.