Question

If two different numbers are taken from the set \left\\{ 0,1,2,...,10 \right\\}; then the probability that their sum as well as absolute difference are both multiple of 4, is:
A.$\dfrac{6}{55}$
B.$\dfrac{12}{55}$
C.$\dfrac{14}{45}$
D.$\dfrac{7}{55}$

Answer 1

Hint: We know that probability is the ratio of total number of favourable cases and the total number of possible outcomes. Here the possible outcomes will be choosing two numbers from the given set. The favourable cases will be selecting those two numbers where the sum and the absolute difference are multiple of 4.

Complete step by step answer:
The probability formula is defined as the probability of an event to happen is equal to the ratio of total number of favourable outcomes and the total number of possible outcomes.
Let us first count the total number of possible outcomes.
We have to choose two different numbers from the set \left\\{ 0,1,2,...,10 \right\\}. In this set we have 11 numbers in total. Out of these 11 numbers we have to select two different numbers.
Therefore we can select 2 different numbers out of 11 in ${}^{11}{{C}_{2}}$ ways.
We know that, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$
Therefore,
${}^{11}{{C}_{2}}=\dfrac{11!}{\left( 11-2 \right)!\times 2!}=\dfrac{11!}{9!\times 2!}=\dfrac{11\times 10\times 9!}{9!\times 1\times 2}=\dfrac{11\times 10}{2}=55$
Hence the total numbers of possible outcomes are 55.
Now we will count the total number of favourable cases.
The favorable cases will be selecting those two numbers where the sum and the absolute difference are multiple of 4.
In the given set the smallest number is 0 and the largest number is 10. So the maximum difference of any two numbers will be 10.
We have to select the numbers in such a way so that the absolute difference is multiple of 4. That means the absolute difference should be either 4 or 8.
Now we will form sets of those numbers who will satisfy the above condition.
Let us take A=\left\\{ 0,4,8 \right\\}.
In the above set if we select any two numbers the absolute difference will be either 4 or 8.
Similarly we can make other sets like this.
B=\left\\{ 1,5,9 \right\\}
C=\left\\{ 2,6,10 \right\\}
D=\left\\{ 3,7 \right\\}
Now if we make more sets the same numbers will be repeated.
Therefore from these above 4 sets if you select any two numbers from the same set the absolute difference will be either 4 or 8.
We have one more condition with this. That is the sum of the two numbers also should be multiple of 4.
Now if we select any two numbers from set A, the sum will be multiple of 4. If we select 0 and 4, the sum is 4, which is multiple of 4. If we select 4 and 8, the sum is 12, which is multiple of 4.
Therefore in set A, we have 3 numbers in total. We can select 2 numbers from 3 numbers in ${}^{3}{{C}_{2}}$ ways.
${}^{3}{{C}_{2}}=\dfrac{3!}{\left( 3-2 \right)!\times 2!}=\dfrac{1\times 2\times 3}{1\times 2}=3$
Similarly if we select any two numbers from set C, the sum will be multiple of 4.
Therefore we can select 2 numbers from 3 numbers of set C in 3 different ways.
But if we select any two numbers from set B or set D, the sum is not a multiple of 4. For example, if we select 5 and 9 from set B, the sum is 14, which is not a multiple of 4.
Therefore set B and C are satisfying only one condition.
But we have to select those numbers which are satisfying both the conditions.
Hence we can only select from set A and set C.
Therefore the total numbers of favourable cases are 3 + 3 = 6
Hence, the probability is $\dfrac{6}{55}$.
Therefore, option (a) is correct.

Note: When we are counting the total number of favourable cases we have to take all the possibilities. Otherwise we will get the wrong answer.If set satisfying one condition and not the second cannot be considered.