A committee of 4 is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have number of boys less than number of girls? A. $$\frac{2}{9}$$ B. $$\frac{4}{9}$$ C. $$\frac{4}{5}$$ D. $$\frac{1}{6}$$

$$ rac{2}{9}$$
$$ rac{4}{9}$$
$$ rac{4}{5}$$
$$ rac{1}{6}$$

The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. In this case, event A is the committee having more girls than boys.

There are $\binom{9}{4} = 126$ ways to choose 4 people from 9 people. There are $\binom{4}{4} = 1$ way to choose a committee with all girls, $\binom{4}{3}\binom{5}{1} = 40$ ways to choose a committee with 3 girls and 1 boy, and $\binom{4}{2}\binom{5}{2} = 60$ ways to choose a committee with 2 girls and 2 boys. Therefore, the probability of the committee having more girls than boys is $\frac{1 + 40 + 60}{126} = \boxed{\frac{101}{126}}$.

Option A is incorrect because it is the probability of the committee having an equal number of boys and girls. Option B is incorrect because it is the probability of the committee having more boys than girls. Option C is incorrect because it is the probability of the committee having at least one girl. Option D is incorrect because it is the probability of the committee having at least one boy.