The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy

The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl?

BGGG

BGBG

BBBG

This is the sameas saying there can be 1boys, 2boys or 3 boys 

so the only thing there cannot be is   BBBB or GGGG

\(\begin{align}\\P(4 boys)& = \frac{10}{20}\times \frac{9}{19}\times \frac{8}{18}\times \frac{7}{17}\\ &=\frac{14}{323}\\ P(4 girls)&=\frac{14}{323}\\ P(4boys \;or\; 4 girls)&=\frac{28}{323}\\ P(not\;all\;the\;same\;s*x)&=1-\frac{28}{323}\\ &=\frac{295}{323} \end{align}\)

See my corrected reply at http://web2.0calc.com/questions/need-asap_1#r3

.

Here's another approach to Melody's answer:

Number of ways to choose a  committee composed of all girls or boys  = 2C(10,4)

Number of possible committees   = C(20,4)

So......the probability that at least one girl or boy is on the committee  =

1   −   2C(10,4)  / C (20,4)    =  1  − 420 / 4845  =  1  − 28.323  = 295 / 393