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Question: An integer is chosen at random from the first 200 positive integers. Find the probability that the i...

An integer is chosen at random from the first 200 positive integers. Find the probability that the integer chosen is divisible by 6 or 8?
A. 14 B. 23 C. 15 D. 0 \begin{gathered} A.{\text{ }}\dfrac{1}{4} \\\ B.{\text{ }}\dfrac{2}{3} \\\ {\text{C}}{\text{. }}\dfrac{1}{5} \\\ D.{\text{ 0}} \\\ \end{gathered}

Explanation

Solution

Hint : Count the total numbers divisible by 6 or 8. Remember to subtract the counting of numbers which are divisible b by both 6 or 8 because otherwise they would have been counted twice.
After counting the total numbers divisible by 6 or 8, find the probability using the formula:
P(E)=number of favourable outcomestotal number of outcomesP\left( E \right) = \dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}

Complete step-by-step answer :
Therefore, 198 is the 33rd number divisible by 6.
So, there are 33 numbers divisible by 6 from 1 to 200.
Total numbers divisible by 8:
The same way we have 200 divisible by 8 and 200 is the 2008=25th\dfrac{{200}}{8} = 25th number divisible by 8.
So, there are 25 numbers divisible by 8 from 1 to 200.
If we want to count numbers divisible by 8 or 6, we add up the numbers we calculated before and subtract the amount of numbers divisible by LCM(6,8)=24LCM\left( {6,8} \right) = 24 because we have counted them twice.
Total numbers divisible by 24:First let's find out how many numbers are divisible by 6 or 8 from 1 to 200.
Total numbers divisible by 6:
We know that 198 is divisible by 6 and the next number divisible by 6 is 204 which is greater than 200.
So, the nth number divisible by 6 is 198.
6n=198 n=1986 n=33 \begin{gathered} \Rightarrow 6n = 198 \\\ \Rightarrow n = \dfrac{{198}}{6} \\\ \Rightarrow n = 33 \\\ \end{gathered}

We have that 192 is divisible by 24 and the next number divisible by 24 is 216 which is greater than 200.
192 is the 8th number divisible by 24 so there are 8 numbers divisible by 24 from 1 to 200.
So, the amount of numbers divisible by 6 or 8 from 1 to 200 is:
  33+258 = 50.\;33 + 25-8{\text{ }} = {\text{ }}50.
Now, probability of an event P(E) is given by:
P(E)=number of favourable outcomestotal number of outcomesP\left( E \right) = \dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}
So, the probability of choosing a number divisible by 6 or 8 from 1 to 200 is:
P(6 or 8)=total numbers divisible by 6 or 8total numbers chosenP\left( {{\text{6 or 8}}} \right) = \dfrac{{{\text{total numbers divisible by 6 or 8}}}}{{{\text{total numbers chosen}}}}
P(6 or 8)=50200=14P\left( {6{\text{ or 8}}} \right) = \dfrac{{50}}{{200}} = \dfrac{1}{4}

Note : The total numbers divisible by 6 or 8 from 1 to 200 can also be counted by the following method:
Total numbers divisible by 6: 2006=33.33=33(rounding of to the smaller nearest integer)\dfrac{{200}}{6} = 33.33 = 33\left( {{\text{rounding of to the smaller nearest integer}}} \right)
Similarly, we can find the total divisible numbers for any integer by this method.