Question

A bag contains 14 identical balls, 4 of which are red, 5 black and 5 white. Six balls are drawn from the bag. find the probability that
(i) 3 are red, (ii) at least two are white.

Answer 1

Solution :

14C6 = 3003 combinations of 14 balls taken 6 at a time.

i)P(3 are red)= (3 red out of 4 red balls) and (3 out of total 10 balls)

$\frac{4c3*10C3}{14C6}$ =$\frac{480}{3003}$ =0.159

ii)

P(at least two white)=1- [P(0 white)+P(1 white)]

that is P(X$\geq 2)$ =1-[P(X=0)+P(x=1)]

= 1- [$\frac{9C6}{14C6}$ + $\frac{5c1*9c5}{14C6}]$

=1-[$\frac{84}{3003}$ + $\frac{5*126}{3003}]$

=1-[$\frac{714}{3003}]$ = $\frac{2289}{3003}$ =0.7622