Question 1200095
<font color=black size=3>
Given table:
<table border = "1" cellpadding = "5">
<tr><td>X(girls)</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td></tr>
<tr><td>P(X)</td><td>0.000</td><td>0.001</td><td>0.006</td><td>0.022</td><td>0.061</td><td>0.122</td><td>0.183</td><td>0.209</td><td>0.183</td><td>0.122</td><td>0.061</td><td>0.022</td><td>0.006</td><td>0.001</td><td>0.000</td></tr>
</table>


Sometimes it's better to transpose the table.
This means each row becomes a column, and vice versa.
We go from a wide table to a tall table.
<table border = "1" cellpadding = "5">
<tr><td>X(girls)</td><td>P(X)</td></tr>
<tr><td>0</td><td>0.000</td></tr>
<tr><td>1</td><td>0.001</td></tr>
<tr><td>2</td><td>0.006</td></tr>
<tr><td>3</td><td>0.022</td></tr>
<tr><td>4</td><td>0.061</td></tr>
<tr><td>5</td><td>0.122</td></tr>
<tr><td>6</td><td>0.183</td></tr>
<tr><td>7</td><td>0.209</td></tr>
<tr><td>8</td><td>0.183</td></tr>
<tr><td>9</td><td>0.122</td></tr>
<tr><td>10</td><td>0.061</td></tr>
<tr><td>11</td><td>0.022</td></tr>
<tr><td>12</td><td>0.006</td></tr>
<tr><td>13</td><td>0.001</td></tr>
<tr><td>14</td><td>0.000</td></tr>
</table>It's up to you which you prefer better.


Use the last four items of the table to say the following:
P(11 or more girls) = P(11 girls) + P(12 girls) + P(13 girls) + P(14 girls)
P(11 or more girls) = 0.022+0.006+0.001+0.000
P(11 or more girls) = <font color=red size=4>0.029</font>



Extra side notes:<ul><li>Luckily we only have four things to add, so it's not too tedious to do this by hand. Spreadsheet software is recommended for larger projects. </li><li>The instructions mention "14 possible values of X", but in reality there are 15 values. Going from X = 1 to X = 14 consists of 14 values. Then X = 0 is the 15th value.</li></ul>
</font>