Question

A box contains 12 chocolates, 3 of which are white chocolate, 4 milk chocolate, and 5 dark chocolate. I am sharing the box with three of my friends. The four of us take turns, each one drawing a chocolate at random from those available in the box. Yes, we're drawing without replacement. We intend to eat the chocolates, not put them back.
Politely, I let my friends draw before me.
For each of the subparts, provide your answer as an unsimplified fractional expression.

a) What is the chance that I draw a dark chocolate?
b) Given that none of my friends draws a dark chocolate, what is the chance that I draw a dark chocolate?
c) What is the chance that all four of us draw milk chocolates?
d) What is the chance that the friend who gets to select first draws a dark chocolate and I draw a dark chocolate too?

Answer 1

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

$\frac{5}{9}$ 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

$\frac{4C4}{12C4} \\=\frac{1}{495}$

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

= $\frac{5}{12} *\frac{7}{11} *\frac{6}{11} +2(\frac{5}{12} *\frac{4}{11} *\frac{7}{11} ) +\frac{5}{12} *\frac{4}{11} *\frac{4}{11} \\= \frac{140+280+80}{1452} \\=\frac{500}{1452} \\=\frac{125}{363}$