A school debate team has 4 girls and 6 boys. A total of 3 of the team members will be chosen to participate in the district debate. What is the probability that 1 girl and 2 boys will be selected?
\(P=\dfrac{\binom{4}{1}\binom{6}{2}}{\binom{6+4}{3}}=\dfrac{4*(6*5/2)}{10*9*8/(3*2)}=1/2\)
The number of all possible 3-person groups
\({10 \choose 3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{6}=120\)
The number of all possible groups consisting of 1 girl and 2 boys
\(4\cdot{6\choose 2}=4\cdot\dfrac{6!}{2!4!}=4\cdot\dfrac{5\cdot6}{2}=60\)
The probability is \(\dfrac{60}{120}=\dfrac{1}{2}\)