Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that

(a) 3 red, 2 blue, and 2 green balls are withdrawn;

(b) at least 2 red balls are withdrawn;

(c) all withdrawn balls are the same color;

(d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Guest Feb 24, 2020

#1**0 **

\(\text{a) $p = \dfrac{\dbinom{12}{3}\dbinom{16}{2}\dbinom{18}{2}}{\dbinom{46}{7}}$}\)

\(\text{b) $P[\text{at least 2 red balls withdrawn}] = 1 - P[\text{0 or 1 red balls withdrawn}]$}=\\ 1 - \dfrac{\dbinom{12}{0}\dbinom{34}{7}}{\dbinom{46}{7}} - \dfrac{\dbinom{12}{1}\dbinom{34}{6}}{\dbinom{46}{7}}\)

\(\text{c) $P[\text{all balls the same color}]$}=\\ \dfrac{\dbinom{12}{7}+\dbinom{16}{7}+\dbinom{18}{7}}{\dbinom{46}{7}}\)

\(\text{d) $P[\text{3 red or 3 blue withdrawn}]$}=\\ \dfrac{\dbinom{12}{3}\dbinom{34}{4} + \dbinom{16}{3}\dbinom{30}{4}}{\dbinom{46}{7}}\)

.Rom Feb 25, 2020