12 boys and 10 girls in which seven students are chosen to go to the blackboard. What is the probability that at least 2 girls are chosen?
1 - (probability of that NOT happening) =
1 - (the probability that all 7 were boys) - (the probability that there was one girl and 6 boys) =
What is the probability that no boys are chosen?
That is the same as all girls are chosen, or
What is the probability that more boys than girls are chosen?
There isn't any shortcut to this:
P(7boys,0girls) + P(6 boys, 1 girl) + P(5 boys, 2 girls) + P(4 boys, 3 girls)
What is the probability that the first three children chosen are boys?
Choose the first three any 3 of the 12 boys C(12,3) ways, and the other 4 any
of the remaining 19 students, C(19,4), so the probability is
Edwin