Question

If two different numbers are taken from the set $\left\{ {0,1,2,3,....,10} \right\}$; then the probability that their sum as well as absolute difference are both multiple of 4, isA.$\dfrac{{12}}{{55}}$B.$\dfrac{{14}}{{45}}$C.$\dfrac{7}{{55}}$D.$\dfrac{6}{{55}}$

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Hint: Here, we are required to find the probability that if two different numbers are taken from the given set, then their sum as well as absolute difference are both multiple of 4. We will find the number of favorable outcomes which are possible i.e. the pair of numbers which satisfy the given conditions. Adding all such cases of the two possible numbers, we will get our favorable outcomes. The given set consists of 11 elements and we have to select 2 elements from this. Hence, we will use combinations to find the total possible outcomes. Dividing the favorable outcomes by the total outcomes, we will get the required probability.

Formula Used:
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where, $n$ represents the total number of terms and $r$represents the number of terms to be selected.
Probability of an event $=$ Number of favourable outcomes $\div$ Total number of outcomes.

We will solve this question by taking various cases from the given set.
According to the first case, let us take the number 0.
Now, let the second number to be chosen be $x$
According to the question, we have to choose the second number such that the sum as well as the absolute difference of the two numbers are both multiple of 4.
This means that $x + 0$ and $x - 0$, both are multiple of 4.
From the given set $\left\{ {0,1,2,3,....,10} \right\}$, the above situation is only possible when the value of $x$ is either 4 or 8
According to the second case, let us take the number 1.
Now, we have to think of the numbers from the given set $\left\{ {0,1,2,3,....,10} \right\}$ with which by adding and subtracting 1 we can get a number divisible by 4.
We will find that no such number exists. Also, this is true for all the odd numbers. Hence, no odd number from the given set paired with any other number will give us a number divisible by 4.
Therefore, we will check only for the even numbers present in the set.
Hence, we will make a table for the possible pairs:

 First Number Second Number Total number of possible cases 0 4 or 8 2 2 6 or 10 2 4 0 or 8(But the case of 0 and 4 is already taken) 1 6 2 or 10 (Here, the case of 2 and 6 is already taken) 1 8 0 or 4 (Here, both the cases are already taken) 0 10 2 or 6 (Again, both the cases are already taken) 0

Hence, total number of possible cases $= 2 + 2 + 1 + 1 = 6$
Hence, total number of favourable outcomes are 6……………………………… $\left( 1 \right)$
Now, the total number of ways of choosing 2 numbers from the given set $\left\{ {0,1,2,3,....,10} \right\}$ where total elements are 11 are: ${}^{11}{C_2} = \dfrac{{11!}}{{2!\left( {11 - 2} \right)!}}$
This is because ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, here, $n$ represents the total number of terms and $r$ represents the number of terms to be selected.
Hence, ${}^{11}{C_2} = \dfrac{{11!}}{{2!\left( {11 - 2} \right)!}} = \dfrac{{11!}}{{2! \times 9!}} = \dfrac{{11 \times 10}}{2} = 11 \times 5 = 55$
Hence, total number of possible outcomes are 55……………………………….. $\left( 2 \right)$
Now, Probability of an event is defined as the number of favourable outcomes divided by the total number of possible outcomes.
Hence, If two different numbers are taken from the set $\left\{ {0,1,2,3,....,10} \right\}$; then the probability that their sum as well as absolute difference are both multiple of 4 $= \dfrac{6}{{55}}$ (from $\left( 1 \right)$and $\left( 2 \right)$)
Therefore, option D is the correct answer.

Note: In mathematics, Probability is used to find how likely an event is to occur. The probability always lies between 0 and 1. Where, if the probability of an event is 0, then this means that it is an impossible event. For example, the probability of getting number 7 while tossing a coin is 0 because a coin will always give a head or a tail and not a number. Similarly, if the probability of an event is 1, then it shows the certainty of a specific event. For example, if there are only blue balls in a bag and hence, the probability of getting blue balls is always 1 because we will get only those balls from the bag as no other colour is present.