A four digit number ( numbered from 0000 to 9999) is said to be lucky if the sum of its first two digits is equal to the sum of its last two digits. If a four digit number is picked at random; then find the probability that it is a lucky number.

(a) Required probability \[=0.067\]
(b) Required probability \[=0.077\]
(c) Required probability \[=0.167\]
(d) Required probability \[=0.177\]

Hint: In this question, we first need to find the number of favourable outcomes that is numbers which have sum of first two digits equal to the sum of last two digits. Then divide the number of favourable outcomes with the total number of possible outcomes which gives the probability.

Complete step-by-step answer:
Let us first express the four digit number according to its place value

\[1000\left( a \right)+100\left( b \right)+10\left( c \right)+d\]

Now, given condition in the question that

\[\Rightarrow a+b=c+d\]

Let us now assume this sum to be some variable n

Now, let us check the possibilities of n.

For \[n=0\] number of possibilities is 1.

For \[n=1\] the number of possibilities are 4.

For \[n=2\] the number of possibilities are 9.

Now, from the above possibilities we can get a generalisation that

For \[n=k\] the number of possibilities will be \[{{\left( k+1 \right)}^{2}}\]

Again for \[n=9\]the number of possibilities will be 100.

Now, for \[n=10\] we get the number of possibilities are 81.

Here, the maximum sum of the digits can be 18.

Now, for the sum to be 10 to the sum 18 the possibilities will be from 81 to 1.

Thus, the total number of lucky numbers are

\[\Rightarrow

{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}......+{{9}^{2}}+{{10}^{2}}+{{9}^{2}}+....+{{1}^{2}}\]

Now, this can be further written as

\[\Rightarrow 2\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}......+{{9}^{2}}+{{10}^{2}} \right)-{{10}^{2}}\]

As we already know that the sum of squares of first n natural numbers is given by the

formula

\[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

Now, by using this formula above equation can be written as

\[\Rightarrow 2\times \dfrac{10\left( 10+1 \right)\left( 2\times 10+1 \right)}{6}-100\]

Now, on further simplification we get,

\[\begin{align}

  & \Rightarrow \dfrac{10\times 11\times 21}{3}-100 \\

 & \Rightarrow 110\times 7-100 \\

 & \Rightarrow 670 \\

\end{align}\]

As we already know from the definition of probability that probability of an event is given by

\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{Number of favourable outcomes}}{\text{Total

number of possible outcomes}}\]

Here, we have from the above conditions that

\[m=670,n=10000\]

Now, by substituting these values in the probability formula we get,

\[\begin{align}

  & \Rightarrow P\left( A \right)=\dfrac{670}{10000} \\

 & \therefore P\left( A \right)=0.067 \\

\end{align}\]

Hence, the correct option is (a).

Note: It is important to note that, from the sum to be 10 the number of possibilities becomes 81 which is square of 9 and again in the same order as they were increased till 10 to the maximum possible sum which is 18 where the possibilities will be 1 same as the possibilities for the sum to be 0.

While calculating it is important to observe that the number of possibilities will be in terms of squares so that we can get a generalisation and can be solved further easily. It is also to be noted that while applying the sum of squares of first n natural numbers we need to subtract squares of 10 as we actually have only once but using the formula we are considering it twice.

Instead of subtracting 100 from the sum of squares of first 10 natural numbers we can also add 100 to the sum of squares of first 9 natural numbers. Both the methods give the same result.