Question

The probability of choosing at random a number that is divisible by 6 or 8 from among 1 to 90 is equal to
A. \[\dfrac{1}{6}\]
B. \[\dfrac{1}{30}\]
C. \[\dfrac{11}{80}\]
D. \[\dfrac{23}{90}\]

Answer 1

Hint: Count the total numbers divisible by 6 or 8. Remember to subtract the counting of numbers which are divisible by both \[6\,\,\text{or}\,\,8\] . Because otherwise they would be counted twice. After counting the total numbers divisible by 6 or 8, find the probability using the formula: \[P(E)=\dfrac{\text{number}\,\,\text{of}\,\,\text{favourable}\,\,\text{outcomes}}{\text{total}\,\,\text{number}\,\,\text{of}\,\,\text{outcomes}}\]

Complete step-by-step answer:
90 is a 15th number divisible by 6
There are 15 numbers divisible by 6 from 1 to 90
Total number divisible by 8:
The same way we have 90 divisible by 8 from 1 to 90 is the \[\dfrac{88}{\text{8}}=\] 11th number divisible by 8
If we want to count numbers divisible by 6 or 8, we have to add up the numbers we calculated before and subtract the amount of numbers divisible by
 \[LCM(6,8)=24\] Because we have counted twice.
Total numbers divisible by 24: first find out how many numbers are divisible by 6 or 8 from 1 to 90
Total numbers divisible by 6:
We know that 90 is divisible by 6 and the next number which is divisible by 6 is 96 which is greater than 90.
So, \[{{n}^{th}}\] number divisible by 6 is 90.
 \[6n=90\]
 \[n=\dfrac{90}{6}\]
 \[n=15\]
We have that 72 is divisible by 24 and the next number divisible by 24 is 96 which is greater than 90.
So, the amount of numbers divisible by 6 or 8 from 1 to 90 is:
 \[15+11-3=23\]
Now, probability of an event \[P(E)\] is given by
\[P(E)=\dfrac{\text{number}\,\,\text{of}\,\,\text{favourable}\,\,\text{outcomes}}{\text{total}\,\,\text{number}\,\,\text{of}\,\,\text{outcomes}}\]
So, the probability of choosing a number divisible by 6 or 8 from 1 to 90.
\[P\left( 6\,\,\text{or}\,\,8 \right)=\dfrac{\text{total numbers}\,\,\text{divisible by}\,\,6\,\,\text{or}\,\,8}{\text{total}\,\,\text{numbers}\,\,\text{chosen}\,\,}\]
\[P\left( 6\,\,\text{or}\,\,8 \right)=\dfrac{23}{90}\]
So, the correct answer is “Option D”.

Note: The total numbers divisible by 6 or 8 from 1 to 90 can also be counted by the following method:
Total numbers divisible by 6 is \[\dfrac{90}{6}=15\] and Total numbers divisible by 8 is \[\dfrac{90}{8}=11.25\] (rounding to the smaller nearest integer). Similarly, we can find the total divisible numbers for any integer by this method.