Question

The set of points $S\in \left( x,y \right)$ are given as x\in \left\\{ 0,1,2,3 \right\\},y\in \left\\{ 0,1,2,3,4 \right\\}. Choose two points with the coordinates belongs to the given sets and get the midpoints of both. So, the probability that the midpoint is chosen belongs to S is
$\left( a \right)\dfrac{21}{95}$
$\left( b \right)\dfrac{42}{95}$
$\left( c \right)\dfrac{36}{95}$
(d) None of these

Answer 1

The selection of ‘r’ things from ‘n’ different things is given as $^{n}{{C}_{r}}.$ Select the number of two points out of the total points formed by the sets x and y. The midpoint of the line joining $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right).$ The midpoints can be integers if ${{x}_{1}}+{{x}_{2}}$ and ${{y}_{1}}+{{y}_{2}}$ are even numbers. The sum of two numbers can be even by adding two even or two odd numbers.

Complete step by step answer:
The set of points $\left( x,y \right)$ are given as x\in \left\\{ 0,1,2,3 \right\\},y\in \left\\{ 0,1,2,3,4 \right\\}.
As, we need to select two points from the sets of ‘x’ and ‘y’. So, one coordinate that is (x, y) can be selected by the given sets of ‘x’ and ‘y’ by $5\times 4=20$ ways that is in total 20 points are possible with the help of given sets of x and y.
Now, we need to select two points by these 20 possible points. So, as we know the selection of ‘r’ different things from ‘n’ different things is done by $^{n}{{C}_{r}}$ ways.
Hence, we can select two points from 20 possible points that is, $^{20}{{C}_{2}}$ ways.
As we are selecting two pairs of points, i.e. $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right),$ so $^{20}{{C}_{2}}$ ways of selection of two pairs of points will consist of the same pairs twice that is if we select $\left( {{x}_{1}},{{y}_{1}} \right)$ first and $\left( {{x}_{2}},{{y}_{2}} \right)$ second as two points. Then this selection will be the same if we select $\left( {{x}_{2}},{{y}_{2}} \right)$ first and $\left( {{x}_{1}},{{y}_{1}} \right)$ second. Hence, $^{20}{{C}_{2}}$ ways consist of the same coordinates twice. So, total different pair of two points possible is $\dfrac{^{20}{{C}_{2}}}{2}.$
Now, there are $\dfrac{^{20}{{C}_{2}}}{2}$ ways of selection of two points. Since, we need to find the mid-point of selected points. As the midpoint of two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right).$
So, it is given that we need to find the probability of the number of mid-points which belong to the given set S: (x, y).
As the formula of mid-points suggests that $\left( {{x}_{1}}+{{x}_{2}} \right)$ and $\left( {{y}_{1}}+{{y}_{2}} \right)$ should be even if $\dfrac{{{x}_{1}}+{{x}_{2}}}{2}$ and $\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$ would be an integer.
So, we have,
x\in \left\\{ 0,1,2,3 \right\\}
y\in \left\\{ 0,1,2,3,4 \right\\}
As we need to select $\left( {{x}_{1}},{{x}_{2}} \right)$ from the set ‘x’ such that $\left( {{x}_{1}}+{{x}_{2}} \right)$ will be even and similarly, we need to select $\left( {{y}_{1}},{{y}_{2}} \right)$ from the set ‘y’ such that $\left( {{y}_{1}}+{{y}_{2}} \right)$ will also be even.
So, the addition of two numbers will be even, if both of them are even or both of them are odd.
So, the integer middle points with $\left( {{x}_{1}},{{x}_{2}} \right)$ both as even numbers will be
$=\left( 0,0 \right),\left( 0,2 \right),\left( 2,0 \right),\left( 2,2 \right)$
= 4 ways
Similarly, integer middle points with $\left( {{y}_{1}}+{{y}_{2}} \right)$ as even numbers = 9.
And integer middle points with $\left( {{y}_{1}},{{y}_{2}} \right)$ as odd numbers = 4.
Hence, the total number of required middle points will be
$=4+4+9+4$
$=21$
So, probability can be given as
$P=\dfrac{\text{Favourable cases}}{\text{Total cases}}=\dfrac{21}{\left( \dfrac{^{20}{{C}_{2}}}{2} \right)}=\dfrac{21}{\left( \dfrac{20\times 19}{2\times 2} \right)}=\dfrac{21}{95}$

So, the correct answer is “Option a”.

Note: One may go wrong if he or she gives the total possible coordinates as $^{20}{{C}_{2}}.$ But we need to select a line with two coordinates, the mid-point of the line segment will not change if the coordinates of two points are exchanged. One may go wrong if he or she uses the formula of $^{n}{{P}_{r}}$ for doing the selection. $^{n}{{P}_{r}}$ gives the arrangement of things. So, be clear with these properties with these kinds of questions.