Question

A box contains a number of marbles with serial number 18 to 38. A marble is picked at random. Find the probability that it is a multiple of 3.
$\left( \text{a} \right)\text{ }\dfrac{3}{5}$
$\left( \text{b} \right)\text{ }\dfrac{7}{20}$
$\left( \text{c} \right)\text{ }\dfrac{3}{4}$
$\left( \text{d} \right)\text{ }\dfrac{1}{3}$

Answer 1

Hint: To solve the above question, we will first find out what probability is and what the formula to calculate the probability of any random event is. Then, we will divide the marbles into three groups. In one group, there will be marbles having the serial number of form 3r. In another group, there will be marbles having the serial number of the form (3r + 1) and in the last group, there will be marbles having the serial number of the form (3r + 2). Then we will choose one marble from the group having 3n serial numbers. This will be favourable outcomes and we will divide it by the total number of marbles to get the probability.

Complete step by step solution:
Before solving the question, we must know what probability is. Probability is defined as the chances of happening in an event. It always lies between 0 and 1. The probability of any random event is denoted by P(E) and is calculated by the formula given below.
$P\left( E \right)=\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$
Now, we have to find the probability that the marble which is picked has a serial number divisible by 3. For divisibility by 3, the number should be of the form 3n. Now, we will make three groups according to the serial number.
In group I, marbles are having the serial number of the form 3r. Thus, this group has serial numbers: 18, 21, 24, 27, 30, 33, 36.
In group II, marbles are having the serial number of the form (3r +1). Thus, this group has serial numbers: 19, 22, 25, 28, 31, 34, 37.
In group III, marbles are having a serial number of the form (3r + 2). Thus, this group has serial numbers: 20, 23, 26, 29, 32, 35, 38.
Now, we have to select a marble from the group I only. There are 7 marbles in this group. So, the favourable outcomes are 7. There are a total 21 marbles in the box. So, the total number of outcomes is 21. Thus, we have,
$P\left( \text{divisible by 3} \right)=\dfrac{7}{21}$
$\Rightarrow P\left( \text{divisible by 3} \right)=\dfrac{1}{3}$
Hence, option (d) is the right answer.

Note: The alternate method to approach this solution is shown below. There are a total 21 marbles which is also the number of total cases. Now, 21 is divisible by 3. So, we can say that there will be $\left( \dfrac{21}{3} \right)$ marbles having serial numbers divisible by 3. This is the number of favourable cases. Thus,
$P\left( \text{divisible by 3} \right)=\dfrac{\left( \dfrac{21}{3} \right)}{21}$
$\Rightarrow P\left( \text{divisible by 3} \right)=\dfrac{21}{3\times 21}$
$\Rightarrow P\left( \text{divisible by 3} \right)=\dfrac{1}{3}$