Answer to Question #120746 in Statistics and Probability for Nimra

Question

A bag contains 5 red 6 green 4 blue and 2 white beads; 6 beads are drawn at random from the bag what is the probability that:

a) 3 are red 2 or green and one is blue?
b) 2 are green 2 are blue and 2 are white?
c) 3 are blue
d) At least 4 are green
e) At least 1 is red bead?

Answer 1

How many beads are there in the bag?

$$5+6+4+2=17$$

a) 3 are red 2 or green and one is blue? 

$$P(3\ red, 2\ green, 1\ blue)=\dfrac{\dbinom{5}{3}\dbinom{6}{2}\dbinom{4}{1}\dbinom{2}{0}}{\dbinom{17}{6}}$$ $$\binom{5}{3}={5!\over 3!(5-3)!}=10$$ $$\binom{6}{2}={6!\over 2!(6-2)!}=15$$ $$\binom{4}{1}={4!\over 1!(4-1)!}=4$$ $$\binom{2}{0}={2!\over 0!(2-0)!}=1$$ $$\binom{17}{6}={17!\over 6!(17-6)!}=12376$$

$$P(3\ red, 2\ green, 1\ blue)={10\cdot 15\cdot4\cdot1\over 12376}={75\over 1547}\approx0.048481$$

b) 2 are green 2 are blue and 2 are white? 

$$P( 2\ green, 2\ blue, 2\ white)=\dfrac{\dbinom{5}{0}\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}{\dbinom{17}{6}}$$ $$\binom{5}{0}={5!\over 0!(5-0)!}=1$$ $$\binom{4}{2}={4!\over 2!(4-2)!}=6$$ $$\binom{2}{2}={2!\over 2!(2-2)!}=1$$

$$P( 2\ green, 2\ blue, 2\ white)={1\cdot 15\cdot6\cdot1\over 12376}={45\over 6188}\approx0.007272$$

c) 3 are blue 

$$P(3\ blue)=\dfrac{\dbinom{4}{3}\dbinom{17-4}{6-3}}{\dbinom{17}{6}}$$ $$\binom{4}{3}={4!\over 3!(4-3)!}=4$$ $$\binom{13}{3}={13!\over 3!(13-10)!}=286$$

$$P(3\ blue)=\dfrac{4\cdot 286}{12376}={11\over 119}\approx0.092437$$

d) At least 4 are green 

$$P(at\ least\ 4\ green)=$$

$$=P(4\ green)+P(5\ green)+P(6\green)=$$

$$=\dfrac{\dbinom{6}{4}\dbinom{17-6}{6-4}}{\dbinom{17}{6}}+\dfrac{\dbinom{6}{5}\dbinom{17-6}{6-5}}{\dbinom{17}{6}}+$$

$$+\dfrac{\dbinom{6}{6}\dbinom{17-6}{6-6}}{\dbinom{17}{6}}$$

$$\binom{6}{4}={6!\over 4!(6-4)!}=15$$ $$\binom{6}{5}={6!\over 5!(6-5)!}=6$$ $$\binom{6}{6}={6!\over 6!(6-6)!}=1$$ $$\binom{11}{2}={11!\over 2!(11-2)!}=55$$ $$\binom{11}{1}={11!\over 1!(11-1)!}=11$$ $$\binom{11}{0}={11!\over 0!(11-0)!}=1$$

$$P(at\ least\ 4\ green)={15\cdot 55\over 12376}+{6\cdot 11\over 12376}+{1\cdot 1\over 12376}=$$

$$={223\over 3094}\approx0.072075$$

e) At least 1 is red bead?

$$P(at\ least\ 1\ red)=1-P(0\ red)=$$

$$=1-\dfrac{\dbinom{5}{0}\dbinom{17-5}{6-0}}{\dbinom{17}{6}}$$

$$\binom{12}{6}={12!\over 6!(12-6)!}=924$$

$$P(at\ least\ 1\ red)=1-{1\cdot 924\over 12376}={409\over 442}\approx0.925339$$