How likely is it that
(a) 4 chocolates are red-wrapped?
$\binom{5}{4}\binom{10}{4}\over\binom{15}{5}$
(b) all chocolates are white-wrapped?
$\frac{10\times9\times8\times7\times6\times5\times4\times3}{15\times14\times13\times12\times11\times10\times9\times8}$
(c) at least one chocolate is red-wrapped?
$\frac{C(10,8)}{C(15,8)}= \frac{45}{6435}$
$P=1-\frac{45}{6435}$
Have I done it correctly? Not sure how to go about it so I just followed the examples from the book. Would appreciate it if anyone could point out my mistakes.
(a) Close. Perhaps it is a typo, but it should be $\dfrac{\dbinom{5}{4}\dbinom{10}{4}}{\dbinom{15}{8}}$
(b) Yes, that is correct.
(c) Correct.