Question

One number is chosen from numbers 1 to 200. Find the probability that it is divisible by 4 or 6?
(a) \[\dfrac{9}{200}\]
(b) \[\dfrac{11}{200}\]
(c) \[\dfrac{91}{200}\]
(d) \[\dfrac{67}{200}\]

Answer 1

Hint: We start solving the problem by finding the total number of terms that were divisible by 4. We then find the total number of terms that were divisible by 6. We then find the number of terms that were divisible by 12 as they were the common terms divisible by both 4 and 6. We then add the total number of terms that were divisible by 4 and 6 and subtract the terms that were divisible by 12 to get the total number of terms that were divisible by 4 or 6. We then find the required probability by using the formula $ \text{probability = }\dfrac{\text{number ways of choosing a number that was divisible by 4 or 6}}{\text{number of ways of choosing a number from the 1 to 200}} $ .

Complete step by step answer:
According to the problem, we are given that one number is chosen from numbers 1 to 200. We need to find the probability that it is divisible by 4 or 6.
Let us first find the total number of terms that were divisible by 4 in the given numbers 1 to 200.
So, the total number of terms divisible by 4 is $ \left[ \dfrac{200}{4} \right]=\left[ 50 \right]=50 $ ---(1). Here $ \left[ . \right] $ is greatest integer function.
Now, let us find the total number of that were divisible by 6 in the given numbers 1 to 200.
So, the total number of terms divisible by 6 is $ \left[ \dfrac{200}{6} \right]=\left[ 33.33 \right]=33 $ ---(2). Here $ \left[ . \right] $ is greatest integer function.
We know that some of the terms which are divided by 4 are also divided by 6. We know that these terms are exactly divisible by their LCM (least common multiple) which is 12.
So, let us find the total number of that were divisible by 12 in the given numbers 1 to 200.
The total number of terms divisible by 12 is $ \left[ \dfrac{200}{12} \right]=\left[ 16.67 \right]=16 $ ---(3). Here $ \left[ . \right] $ is greatest integer function.
So, we subtract these common terms as they were included in both the number of terms that were divisible by 4 or 6.
Now, let us find the total number of terms that were divisible by 4 or 6 which is $ 50+33-16=67 $ .
Let us now find the probability that the chosen number is divisible by 4 or 6.
We know that $ \text{probability = }\dfrac{\text{Number of ways of choosing a number that were divisible by 4 or 6}}{\text{number of ways of choosing a number from the 1 to 200}} $ .
So, we get $ \text{probability = }\dfrac{67}{200} $ .
So, we have found the required probability as $ \dfrac{67}{200} $ .
∴ The correct option for the given problem is (d).

Note:
 Whenever we get this type of problem, we first try to start by finding the total number of favorable cases. We should forget to subtract the terms that were commonly divisible by 4 or 6 as it is the most common mistake by students. We can also write the terms that were divisible by 4 or 6 and count them manually instead of making use of the greatest integer function. Similarly, we cannot expect problems to find the probability that the chosen number is divisible by 4 and 6.