Answer:
Step-by-step explanation:
Given that a box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate.
a) the chance that I draw a dark chocolate
b) Given that none of my friends draws a dark chocolate, the chance that I draw a dark chocolate is
since now all 5 dark chocolates would be there and total changed to 9 because 3 friends took 1 each.
c) All four drawn milk chocolates would be
![\frac{4C4}{12C4} \\=\frac{1}{495}](https://tex.z-dn.net/?f=%5Cfrac%7B4C4%7D%7B12C4%7D%20%5C%5C%3D%5Cfrac%7B1%7D%7B495%7D)
d) the chance that the friend who gets to select first draws a dark chocolate and I draw a dark chocolate too
=P(first friend draws a dark chocolate) * P(I draw a dark chocolate)
When it comes to me chances are either there are 4 dark, or 3 dark, or 1 dark depending upon the other two friends drew black or not
= ![\frac{5}{12} *\frac{7}{11} *\frac{6}{11} +2(\frac{5}{12} *\frac{4}{11} *\frac{7}{11} ) +\frac{5}{12} *\frac{4}{11} *\frac{4}{11} \\= \frac{140+280+80}{1452} \\=\frac{500}{1452} \\=\frac{125}{363}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B12%7D%20%2A%5Cfrac%7B7%7D%7B11%7D%20%2A%5Cfrac%7B6%7D%7B11%7D%20%2B2%28%5Cfrac%7B5%7D%7B12%7D%20%2A%5Cfrac%7B4%7D%7B11%7D%20%2A%5Cfrac%7B7%7D%7B11%7D%20%29%20%2B%5Cfrac%7B5%7D%7B12%7D%20%2A%5Cfrac%7B4%7D%7B11%7D%20%2A%5Cfrac%7B4%7D%7B11%7D%20%5C%5C%3D%20%5Cfrac%7B140%2B280%2B80%7D%7B1452%7D%20%5C%5C%3D%5Cfrac%7B500%7D%7B1452%7D%20%5C%5C%3D%5Cfrac%7B125%7D%7B363%7D)