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

(i) There are 19 balls of which 6 are white, 4 are black and 9 are black.

Number of ways 3 balls can be drawn is $C^3_{19}$ .

Number of ways 2 white balls can be drawn is $C^2_6$.

Number of ways 1 not white ball can be drawn is $C^1_{13}.$

$P(2\; white)=\frac{C^2_6*C^1_{13}}{C^3_{19}}=\frac{6!*13!*3!*16!}{2!*4!*19!}=\frac{65}{323}\approx0.2012.$

(ii) Number of ways 1 white ball can be drawn is $C^1_6$.

Number of ways 1 red ball can be drawn is $C^1_4$ .

Number of ways 1 black ball can be drawn is $C^1_9$.

$P(one\; of\; each\; color)=\frac{C^1_6*C^1_4*C^1_9}{C^3_{19}}=\frac{6!*4!*9!*3!*16!}{1!*5!*1!*3!*1!*8!*19!}=\frac{72}{323}\approx0.2229.$

(iii) Number of ways 3 not red balls can be drawn is $C^3_{(19-4)}=C^3_{15}$

$P(none\; red)=\frac{C^3_{15}}{C^3_{19}}=\frac{15!*3!*16!}{3!*12!*19!}=\frac{91}{323}\approx0.2817.$

(iv) Number of ways 3 not white balls can be drawn is $C^3_{(19-6)}=C^3_{13}$ .$P(at\; least\; 1\; white)=1-P(no\; white)=1-\frac{C^3_{13}}{C^3_{19}}=1-\frac{13!*3!*16!}{3!*10!*19!}=1-\frac{286}{969}\approx0.7049.$