Question

A number of six digits is written down at random. Probability that sum of digits of the number is even is:
A) \[\dfrac{1}{2}\]
B) \[\dfrac{3}{8}\]
C) \[\dfrac{3}{7}\]
D) None of these

Answer 1

Hint:
Here we will firstly write the possible numbers that can be written. Then we will find the probability of selecting the even number and selection of odd numbers among the numbers available. Then we will form the possible conditions to get the sum of digits of the number as even. Then we will form the equation and solve it to get the Probability that the sum of digits of the number is even.

Complete step by step solution:
Let E be the event of the number selected being an even number. Similarly let O be the event of the number selected being an odd number.
Total number of digits available is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Therefore, we get
Number of digits available \[ = 10\]
Now we will write the number of even digits available and number of odd digits available. Therefore, we get
Number of even digits \[ = 5\]
Number of odd digits \[ = 5\]
Now we will find the probability of selection of the even digit. Therefore, we get
 \[P\left( E \right) = \dfrac{5}{{10}} = \dfrac{1}{2}\]
Now we will find the probability of selection of the odd digit. Therefore, we get
 \[P\left( O \right) = \dfrac{5}{{10}} = \dfrac{1}{2}\]
Now we will find all the possible combinations of the numbers such that the sum of digits of the number is even. We know that the sum of the two even numbers is equal to even and the sum of two odd numbers is also equal to even but the sum of one odd and one even number is always equal to odd.
Therefore, all the possible cases for sum of digits of the number is even are
 \[\begin{array}{l}E + E + E + E + E + E = E\\E + E + E + E + O + O = E\\O + O + E + E + E + E = E\\O + O + O + O + O + O = E\end{array}\]
Now we will find the probability by using these conditions. Therefore, we get
Probability that sum of the digits is even \[ = {}^6{C_6}{p^6}{q^0} + {}^6{C_4}{p^4}{q^2} + {}^6{C_2}{p^2}{q^4} + {}^6{C_0}{p^0}{q^6}\]
Now by solving this we will get the probability that sum of the digits is even, we get
 \[ \Rightarrow \] Probability that sum of the digits is even \[ = \left( {1 \times {{\left( {\dfrac{1}{2}} \right)}^6} \times {{\left( {\dfrac{1}{2}} \right)}^0}} \right) + \left( {\dfrac{{6 \times 5 \times 4 \times 3}}{{4 \times 3 \times 2 \times 1}} \times {{\left( {\dfrac{1}{2}} \right)}^4} \times {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \left( {\dfrac{{6 \times 5}}{{2 \times 1}} \times {{\left( {\dfrac{1}{2}} \right)}^2} \times {{\left( {\dfrac{1}{2}} \right)}^4}} \right) + \left( {1 \times {{\left( {\dfrac{1}{2}} \right)}^0} \times {{\left( {\dfrac{1}{2}} \right)}^6}} \right)\]
 \[ \Rightarrow \] Probability that sum of the digits is even \[ = \left( {{{\left( {\dfrac{1}{2}} \right)}^6}} \right) + \left( {15 \times {{\left( {\dfrac{1}{2}} \right)}^6}} \right) + \left( {15 \times {{\left( {\dfrac{1}{2}} \right)}^6}} \right) + \left( {{{\left( {\dfrac{1}{2}} \right)}^6}} \right)\]
Now taking \[{\left( {\dfrac{1}{2}} \right)^6}\] common from all the terms, we get
 \[ \Rightarrow \] Probability that sum of the digits is even \[ = {\left( {\dfrac{1}{2}} \right)^6}\left( {1 + 15 + 15 + 1} \right)\]
Simplifying the expression, we get
 \[ \Rightarrow \] Probability that sum of the digits is even \[ = \left( {\dfrac{1}{{64}}} \right)\left( {32} \right)\]
Multiplying the terms, we get
 \[ \Rightarrow \] Probability that sum of the digits is even \[ = \dfrac{1}{2}\]
Hence, the Probability that the sum of digits of the number is even is \[\dfrac{1}{2}\].

So, option A is the correct option.

Note:
Probability is the branch of mathematics which gives the possibility of the event occurrence and is equal to the ratio of the number of favorable outcomes to the total number of outcomes. Probability generally lies between the values of 0 to 1. Similarly in the above question, we got the probability in the range of 0 and 1. Probability can never be above the value of 1 or less than 0. So if we got the value of probability greater than 1 then we need to check out the calculation to remove the error for calculating the probability.