Question

In a class of 40 students, ask them to write a 3 – digit number. Choose any student at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Answer 1

Hint: Calculate the probability of choosing any one student in a class of 40 students. Calculate the probability of getting a three digit number that is divisible by 3. Multiply the probability of choosing a student with the probability of choosing a three digit number which is divisible by 3 to get the probability of choosing a student who wrote a three digit number which is divisible by 3.

Complete step-by-step solution -

We have to calculate the probability of choosing a student who wrote a three digit number which is divisible by 3.
We will firstly evaluate the probability of choosing a student out of a class of 40 students.
We know that probability of any event is defined as the ratio of number of favourable outcomes to the total number of possible outcomes.
Let us denote the event of choosing a student by A.
Thus, we have $P\left( A \right)=\dfrac{1}{40}$.
We will now evaluate the probability of choosing a three digit number which is divisible by 3 out of all the three digit numbers. Let us denote this event by B.
We will firstly count the total number of possible three digit numbers. We observe that in any three digit number, 9 digits are possible at hundreds place, 10 digits are possible at tens place and 10 digits are possible at units place.
So, the total number of three digit numbers possible $=9\times 10\times 10=900$.
We will now count the number of three digit numbers which are divisible by 3. We observe that the first three digit number divisible by 3 is 102 and the last 3 digit number divisible by 3 is 999.
We also observe that all the three digit numbers divisible by 3 are in AP with first term 102 and common difference 3. The last term of this AP is 999. We have to calculate the number of terms in this AP. Let’s assume that this AP has n terms. So, the ${{n}^{th}}$ term of this AP is 999.
We know that ${{n}^{th}}$ term of the AP whose first term is a and common difference is d is given by ${{a}_{n}}=a+\left( n-1 \right)d$.
Substituting $a=102,d=3,{{a}_{n}}=999$ in the above expression, we have $999=102+\left( n-1 \right)3$.
Further simplifying the equation, we have $n-1=\dfrac{999-102}{3}=299$.
Thus, we have $n=299+1=300$.
Thus, there are $n=300$ three digit numbers which are divisible by 3.
So, the probability of choosing a three digit number which is divisible by 3 out of the set of all three digit numbers $=P\left( B \right)=\dfrac{300}{900}=\dfrac{1}{3}$.
We will now calculate the probability of choosing a student who writes a three digit number divisible by 3. This is given by $P\left( A\cap B \right)$.
As A and B are independent events, we have $P\left( A\cap B \right)=P\left( A \right)P\left( B \right)$.
Thus, we have $P\left( A\cap B \right)=P\left( A \right)P\left( B \right)=\dfrac{1}{40}\times \dfrac{1}{3}=\dfrac{1}{120}$.
Hence, the probability of choosing a student who writes a three digit number divisible by 3 is $\dfrac{1}{120}$.

Note: Two events are said to be independent when the occurrence or non occurrence of one event doesn’t affect the occurrence or non occurrence of another event. Probability of intersection of two independent events is given by product of probability of each of the events. While counting the three digit numbers, we must consider the repetition of digits. Also, one can’t place 0 at the hundreds place of a three digit number.