There are 4 boys and 3 girls. What is the probability the boys and girls sit alternately?

I'll help you get started on this. For them to sit alternately the first one in line must be a boy. There are 7 children and 4 are boys, so the probability that a boy is first is 4/7. The second person in line must be a girl, and if a boy was first then there are 3 girls out of 6 children remaining, so the probability of the 2nd child being a girl is 3/6. Hence the probability that the first child is a boy and the second child is a girl is 4/7 x 3/6. Now continue with this line of thinking to fill the remaining 5 slots. Post back your answer and we'll check it for you.