A team of 5 is chosen at random from a club of 6 boys and 4 girls. What is the probability that the team includes at least one girl?

'At least one' calls for the complement. The teams number $C(10,5)=252$, and a team with no girl at all must take all 5 members from the 6 boys, which can be done in $C(6,5)=6$ ways. So at least one girl appears in 252 − 6 = 246 teams, a probability of 246/252 = 41/42. The value 1/42 = 6/252 is the all-boy case left un-subtracted. The value 5/21 = 60/252 is the probability of EXACTLY one girl, from $C(4,1)\times C(6,4)=4\times 15=60$, which is a much smaller event than 'at least one'. The value 2/5 = 4/10 is the fraction of the club that is female, which would be the answer if a single member were being picked rather than a team of five. With only 6 boys available for 5 places, an all-boy team is very hard to avoid, so the answer being so near 1 is exactly what should be expected.