Question

Six white balls and four black balls which are indistinguishable apart from color are placed in a bag. If six balls are taken from the bag, find the probability of their being three white and three black?

Answer 1

For 3 white balls (3 from 6) the number of combinations is

$\displaystyle C(6,3)=\dfrac{6!}{3!3!}=\dfrac{3!\cdot4\cdot5\cdot6}{3!3!}=$ $\dfrac{4\cdot5\cdot6}{3!}=\dfrac{4\cdot5\cdot6}{1\cdot2\cdot3}=4\cdot5=20$

For 3 black balls (3 from 4) the number of combinations is

$\displaystyle C(4,3)=\dfrac{4!}{3!1!}=\dfrac{3!\cdot4}{3!1!}=4$

For 3 white and 3 black balls use multiplication principle

$m=\displaystyle C(6,3)\cdot\displaystyle C(4,3)=20\cdot4=80$

For 6 taken balls (3+3 from 6+4) the number of combinations is

$n=\displaystyle C(10,6)=\dfrac{10!}{6!4!}=$ $\dfrac{6!\cdot7\cdot8\cdot9\cdot10}{6!4!}=$ $\dfrac{7\cdot8\cdot9\cdot10}{1\cdot2\cdot3\cdot4}=$ $\dfrac{7\cdot9\cdot10}{3}=$

$7\cdot3\cdot10=210$

The probability is

$P=\dfrac{m}{n}=\dfrac{C(6,3)\cdot C(4,3)}{C(10,6)}=\dfrac{80}{210}=\dfrac{8}{21}$