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One integer is chosen out of \[1,2,...,100\]. What is the probability that it is neither divisible by $4$ nor by $6$.
A. $0.59$
B. $0.67$
C. $0.41$
D. $0.33$

Answer
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Hint: To find the probability that an integer is neither divisible by \[4\] nor by $6$, we have to find the total number of integers from 1,2,….100 which are divisible by 4, divisible by 6 and divisible by both 4 and 6. Then using the formula for set theory we calculate the number of integers that are divisible by 4 or by 6. Then we subtract the number of integers divisible by 4 or by 6 from the total number of integers.
* \[n(A \cup B) = n(A) + n(B) - n(A \cap B)\]
* Probability of an event is the number of favorable outcomes divided by the total number of outcomes.

Complete step-by-step answer:
Let us assume the number of integers divisible by 4 be denoted by \[n(A)\]
Number of integers divisible by 6 be denoted by \[n(B)\]
Number of integers divisible by both 4 and 6 will be denoted by \[n(A \cap B)\]
Therefore, the number of integers divisible by either 4 or by 6 will be denoted by \[n(A \cup B)\]
Find the total numbers which are divisible by $4$.
Numbers from $1,2,...,100$ which are divisible by $4$ are written as multiples of 4 which are less than or equal to 100.
$1 \times 4 = 4,2 \times 4 = 8,...,25 \times 4 = 100$.
$\therefore $ Total numbers which are divisible by \[4 = 25\]
\[ \Rightarrow n(A) = 25\]
Find the total numbers which are divisible by $6.$
Numbers from $1,2,...,100$ which are divisible by $6$ are written as multiples of 6 which are less than or equal to 100.
$1 \times 6 = 6,2 \times 6 = 12,...,16 \times 6 = 96$
$\therefore $ Total numbers which are divisible by \[6 = 16\]
\[ \Rightarrow n(B) = 16\]
Find the total numbers which are divisible by both \[4, 6\].
Here we take LCM of \[4, 6\] as \[12\]. So, the numbers which are divisible by both 4 and 6 are divisible by 12.
Total numbers which are divisible by 12 are written as multiples of 12 which are less than or equal to 100.
$1 \times 12 = 12,2 \times 12 = 24,...,8 \times 12 = 96$
$\therefore $ Total numbers which are divisible by \[12 = 8\]
\[ \Rightarrow n(A \cap B) = 8\]
Substitute the values in the formula \[n(A \cup B) = n(A) + n(B) - n(A \cap B)\]
\[
   \Rightarrow n(A \cup B) = 25 + 16 - 8 \\
   \Rightarrow n(A \cup B) = 41 - 8 = 33 \\
 \]
So, the number of integers divisible by either 4 or by 6 is \[n(A \cup B) = 33\]
Then, number of integers which are neither divisible by 4 nor by 6 is given by total number of integers \[ - n(A \cup B)\]
\[ \Rightarrow 100 - n(A \cup B) = 100 - 33 = 67\]
Now calculate the probability that number is neither divisible by 4 nor by 6
Since, we know Probability of an event is the number of favorable outcomes divided by the total number of outcomes.
Here, the number of favorable outcomes is 67 and the total number of outcomes is 100.
Probability ${\text{ = }}\dfrac{{67}}{{100}} = 0.67.$

So, the correct answer is “Option B”.

Note: Students are likely to miss out the number of integers that are divisible by both 4 and 6, and they tend to subtract both number of integers divisible by 4 and number of integers divisible by 6 from total number of integers and find the probability which is wrong, there are some numbers which are divisible by both 4 and 6 together.