JEE Probability 1

case-1

x or y divisible by 6,

Possibilities =

$\left\lfloor \frac { 100 }{ 6 } \right\rfloor (100)\quad +\quad \left\lfloor \frac { 100 }{ 6 } \right\rfloor (100)\quad -\quad { \quad \left\lfloor \frac { 100 }{ 6 } \right\rfloor }^{ 2 }\quad =\quad 2944$

(using inclusion exclusion principle)

case -2

x divisible by 2 but not by 6 and y divisible by 3 but not by 6 and vice versa

$2(\left\lfloor \frac { 100 }{ 2 } \right\rfloor -\left\lfloor \frac { 100 }{ 6 } \right\rfloor )(\left\lfloor \frac { 100 }{ 3 } \right\rfloor -\left\lfloor \frac { 100 }{ 6 } \right\rfloor )=1156$

( it is not possible for something to be divisible by 2 and 3 simultaneously unless it is also divisible by 6, so we dont need to worry about any overlaps)

total = 4100

and total number of ways = 100*100 = 10000

so probability is 0.41

hence 1000 times it is 410 (answer)

Let $x=6k_1+r_1$ and $y=6k_2+r_2$ where $k_1,k_2$ are non-negative integers and $r_1,r_2\in \{0,1,2,3,4,5\}$ .

$\bullet$ Case 1: $r_1=0,$ In this case, $r_2$ can be {0,1,2,3,4,5} . Number of such values of $x=16$ and number of such values of $y=100$ . * 1600 favourable pairs! *

$\bullet$ Case 2: $r_1=1,$ In this case, $r_2$ can be {0} . Number of such values of $x=17$ and number of such values of $y=16$ . 272 favourable pairs!

$\bullet$ Case 3: $r_1=2,$ In this case, $r_2$ can be {0,3} . Number of such values of $x=17$ and number of such values of $y=33$ . 561 favourable pairs!

$\bullet$ Case 4: $r_1=3,$ In this case, $r_2$ can be {0,2,4} . Number of such values of $x=17$ and number of such values of $y=50$ . 850 favourable pairs!

$\bullet$ Case 5: $r_1=4,$ In this case, $r_2$ can be {0,3} . Number of such values of $x=17$ and number of such values of $y=33$ . 561 favourable pairs!

$\bullet$ Case 6: $r_1=5,$ In this case, $r_2$ can be {0} . Number of such values of $x=16$ and number of such values of $y=16$ . 256 favourable pairs!

Total favourable cases $=1600+272+561+850+561+256=4100$

Probability $=\dfrac{4100}{100\times 100}=0.41$

$\eta =\lfloor 1000\times 0.41\rfloor=\boxed{410}$

I'm not sure if someone covered it this way but one very easy way to look at it is this - If I choose just one multiple of $2$ and one multiple of $3$ (without worrying whether it is a multiple of $6$ or not) I will end up covering almost all cases apart from the ones where one is a multiple of $6$ and the other isn't a multiple of $2$ and $3$ .

Going by that, we get $(50\times 33)\times 2$ (where $\times 2\$ is for arrangement) as the first case, but the important thing to note here is that some cases will duplicate, namely the ones where both are multiples of $6$ . So in totality, the first case becomes $(50\times 33)\times 2 -256$ .

The next case is pretty straightforward - $(16\times 33)\times 2$ .

This brings the total to $4100$ . The rest is simple.

The numbers can be classified as

Let $\xi = \text{gcd}(x, 6)$ and $\eta = \text{gcd}(y, 6)$ . The distribution of $\xi$ is as follows: $f(\xi = 6) = 16;\ \ f(\xi = 3) = 17;\ \ f(\xi = 2) = 34;\ \ f(\xi = 1) = 33,$ and similar for $\eta$ .

Now $6 | xy$ iff $6 | \xi\eta$ . Its distribution is as follows:

$\begin{array}{r|cccc} & \xi = 6 & \xi = 3 & \xi = 2 & \xi = 1 \\ \hline \eta = 6 & Y & Y & Y & Y \\ \eta = 3 & Y & N & Y & N \\ \eta = 2 & Y & Y & N & N \\ \eta = 1 & Y & N & N & N \end{array}$

We must add the frequencies for all "Y" entries in the table. This is $2\cdot 100\cdot 16 - 16^2 + 2\cdot 17\cdot 34 = 4100.$ Therefore the desired probability is $0.410$ and we submit $\boxed{410}$ for an answer.

{x y total} indicates no of values of x ,y and total possible outcomes xy. {16 100 1600 } ( x being multiple of 6 and y any from 1 to 100) {84 16 1344 }(y being multiple of 6 and x be any from 1 to 100 except multiples of 16) {34 17 578} (x being multiples of 2 which are not multiples of 6 and y being multiple of 3 which are not multiples of 2) {17 34 578}(interchanging x and y in above ) total 4100 probability :- 4100/10000 1000 *probability =410