A number is selected at random from the first 50 natural numbers. The probability that the selected number is a multiple of 3 or 4 is: A. $\dfrac{12}{25}$ B. $\dfrac{14}{25}$ C. $\dfrac{14}{50}$ D. $\dfrac{8}{25}$
ANSWER
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Hint: Count all the numbers that are divisible by 3 first, then count all the numbers divisible by 4. Then count the total numbers and subtract the ones that have been added extra. Then use the formula of probability and get the desired result.
Complete step by step answer: In the question, we have been asked to find the probability of selecting a number from the list of first 50 natural numbers that are divisible by either 3 or 4. Let us know a bit about probability first. Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur or how likely a proposition is to be true. Probability is a number between 0 and 1, where roughly speaking, 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely is the possibility that it will occur. A simple example of the same is, the tossing of a fair (unbiased) coin. Since the coin is fair, the outcomes are both equally probable, ‘heads’ or ‘tails’. The probability of ‘heads’ is equal to the probability of ‘tails’. And since no other outcome is possible, the possibility of either ‘heads’ or ‘tails’ is $\dfrac{1}{2}$. (which can also be written as 0.5 or 50%) It is represented by the formula, $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$, where, $P\left( A \right)$ represents the probability of event $A$, $n\left( A \right)$ represents the number of outcomes that support event $A$, and $n\left( S \right)$ represents the total number of outcomes which happen which counts the favourable and unfavourable ones too. Now, the numbers that are divisible by 3 are the multiples of 3 which will continue till 48, and they are listed as:$\left\{ 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48 \right\}$, so the total count is 16. Now, the numbers that are divisible by 4 are the multiples of 4 which will continue till 48, and they are listed as : $\left\{ 4,8,12,16,20,24,28,32,36,40,44,48 \right\}$, so the total count is 12. In the above two sets we can see that there are 4 elements that are added extra such as $\left\{ 12,24,36,48 \right\}$. So, the total favourable outcomes will be $16+12-4=24$. Hence, the probability of the event will be $\dfrac{24}{50}$ as the total number of outcomes is 50. Therefore, the answer is $\dfrac{24}{50}=\dfrac{12}{25}$, so option A is the correct answer.
Note: Most of the time the students forget to consider those 4 extra elements, which leads them to the wrong option B instead of option A, making their answer incorrect. Here we had a sample space of 50 natural numbers. So, the total outcomes were 50.