Question 1157959
<font color=black size=3>
This is a great question. Consider we have a 4x4 grid as shown below
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>A1</td><td>A2</td><td>A3</td></tr>
<tr><td>A4</td><td>A5</td><td>A6</td><td>A7</td></tr>
<tr><td>A8</td><td>A9</td><td>A10</td><td>A11</td></tr>
<tr><td>A12</td><td>A13</td><td>A14</td><td>A15</td></tr>
</table>


There are 16 items total. One of which is X. The other 15 are A1 through A15. The probability of randomly selecting X is 1/16


Now consider this grid
<table border = "1" cellpadding = "5">
<tr><td>Y</td><td>B1</td><td>B2</td></tr>
<tr><td>B3</td><td>B4</td><td>B5</td></tr>
<tr><td>B6</td><td>B7</td><td>B8</td></tr>
</table>


We have 9 items. One is Y, the other 8 are B1 through B8. The probability of randomly selecting Y is 1/9


I claim that the probability of selecting both X and Y at the same time, again completely randomly, is 1/144 which is the result of multiplying 1/16 and 1/9. Both events are independent as neither grid interacts with one another. 


This is not obvious why this is the case. But let's say we arranged the items in the first grid to be along the left side of a very large table (16 rows) and the items of the second grid to be along the top of the large table (9 columns). We have this blank table
<table border = "1" cellpadding = "5">
<tr><td></td><td>Y</td><td>B1</td><td>B2</td><td>B3</td><td>B4</td><td>B5</td><td>B6</td><td>B7</td><td>B8</td></tr>
<tr><td>X</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A1</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A2</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A3</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A4</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A5</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A6</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A7</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A8</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A9</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A10</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A11</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A12</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A13</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A14</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A15</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
</table>



I'll fill out part of the table. The first row shows
<font color=blue>XY, XB1, XB2, ... , XB8</font>
<table border = "1" cellpadding = "5">
<tr><td></td><td>Y</td><td>B1</td><td>B2</td><td>B3</td><td>B4</td><td>B5</td><td>B6</td><td>B7</td><td>B8</td></tr>
<tr><td>X</td><td><font color=blue>XY</font></td><td><font color=blue>XB1</font></td><td><font color=blue>XB2</font></td><td><font color=blue>XB3</font></td><td><font color=blue>XB4</font></td><td><font color=blue>XB5</font></td><td><font color=blue>XB6</font></td><td><font color=blue>XB7</font></td><td><font color=blue>XB8</font></td></tr>
<tr><td>A1</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A2</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A3</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A4</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A5</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A6</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A7</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A8</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A9</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A10</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A11</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A12</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A13</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A14</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
<tr><td>A15</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr>
</table>



Notation:
XY = we selected X and Y
XB1 = we selected X and B1
XB2 = we selected X and B2
etc etc


The rest of the table is filled out the same way. You just use the headers as shown. Anyways, note how there are 16 rows and 9 columns giving 16*9 = 144 items total in this massive table. Ask yourself: how many items are XY? That would be one only. This shows the probability of getting XY randomly is 1/144.


------------------------------------------------


As you can see, this can be thought of as an area problem. Areas of rectangles are found by multiplying the length and width. Squares are a special type of rectangle. Consider a square of area 1. This represents 100% probability.



Draw a square with area 1. Mark the upper half as a blue rectangle. If we throw a dart, then the probability of landing in the blue shaded region is 1/2. This is because 1/2 of the area is blue, and the total area is 1, so (1/2)/1 = 1/2.
<img width="25%" src = "https://i.imgur.com/r3KifXu.png">
Square divided into 2 equal pieces


Draw another square of area 1. Divide the rectangle into 3 vertical equal slices (think of a candy bar being broken up perhaps).
<img width="25%" src = "https://i.imgur.com/kykd4p4.png">
Square divided into 3 equal pieces. The probability of landing in the red region is 1/3.


If we overlap the two figures, then we'll have the blue and red regions combine to form the purple area
<img width="25%" src = "https://i.imgur.com/giWhNnC.png">
Square divided into 6 equal pieces


Ask yourself this: what is the probability of landing in BOTH the upper blue area and the red area at the same time? That would be 1/6. The area of the blue rectangle is (1/2)*1 = 1/2. The area of the red rectangle is 1*(1/3) = 1/3. So the purple region has area (1/2)*(1/3) = 1/6


Or you can think of it like this: Start with a red rectangle (area 1/3) and cut it in half to get the purple rectangle (area 1/6). A slight variation: Start with a blue rectangle (area 1/2) and cut that into thirds to get the purple piece (area 1/6)
</font>