Answer to Question #105001 in Statistics and Probability for Riya

Question

An urn contains 6 white, 4 red and 9 black balls. If 3 balls are drawn at random, find
the probability that: (i) two of the balls drawn are white, (ii) one is of each colour,
(iii) none is red, (iv) at least one is white.

Answer 1

There are 19 balls of which 6 are white, 4 are red and 9 are black. Number of ways N 3 balls can be drawn is 

$$\binom{19}{3}={19! \over 3!(19-3)!}={19(18)(17) \over 1(2)(3)}=969$$

(i) Find the probability that two of the balls drawn are white

$$P(two\ white)={\binom{6}{2}\binom{19-6}{1} \over 969}={{6! \over 2!(6-2)!}(13) \over 969}={195 \over 969}={65 \over 323}$$

(ii) Find the probability that one is of each colour

$$P(each\ color)={\binom{6}{1}\binom{4}{1}\binom{9}{1} \over 969}={6(4)(9) \over 969}={216 \over 969}={72 \over 323}$$

(iii) Find the probability that none is red

Out of 19 balls 4 are red and 15 are not red.

$$P(no \ red)={\binom{19-4}{3} \over 969}={15! \over 3!(15-3)!}\cdot{1 \over 969}=$$ $$={15(14)(13) \over 1(2)(3)(969)}={455\over 969}$$

(iv) Find the probability at least one is white.

Out of 19 balls 6 are white and 13 are not white.

$$P(no \ white)={\binom{19-6}{3} \over 969}={13! \over 3!(13-3)!}\cdot{1 \over 969}=$$ $$={13(12)(11) \over 1(2)(3)(969)}={286\over 969}$$

$$P(at\ least\ one \ white)=1-P(no \ white)=$$ $$=1-{286\over 969}={683\over 969}$$