5 children so you have 2^5=32 possibilities to "assign" genders P(3 girls): how many possibilities are there to "assign" the 3 girl-genders to the 5 children? the first girl has 5 possibilities then the next 4, 3 -> 5*4*3=60 but these possibilities include orders of assigned genders, while children 1-5 might differ the gender "girl" is always the same so we have the remove the orderings of the 3 girl-gender assignments which is 3*2*1=6 if we divide 60/6 we get 10 possibilities to have 3 girls, so what is the resulting chance? the 10 possibilities divided by the total 32 possibilities: 10/32=5/16=P(3 girls)=P(2 boys)
this is a bit of lengthy way of saying "use the binomial coefficient" equation/explaining it a bit which is (n!)/(k!(n-k)!) with n=5, k=3: 5*4*3*2*1/((3*2*1)*(2*1))= 5*4*3*2/(3*2*2)= 5*4*3*2/(3*4)= 5*2= 10 possibilities again
P(boys=0) is the easy case: simply multiply the chance of getting a girl 5 times: (1/2)^5=1/32 P(boys=1)= again the binomial coefficient with n=5 and k=1: 5*4*3*2*1/((1)*(4*3*2*1))= 5*4*3*2/(4*3*2)= 5 possibilities so the P(boys=0)=1 possibility + P(boys=1)=5 possibilities totals to 6 possibilities again the chance is the 6 possibilities divided by all 32 possibilities: 6/32=3/16
P(alternate gender starting with boy): when thinking about the possibilities then there is only a single way to build that order: bgbgb, so one possibility knowing there is only one way we already know P(alternate...)=1/32 by again dividing by the total amount of possibilities the alternative way would be to multiply P(boy)*P(girl)*P(boy)*P(girl)*P(boy)=(1/2)^5= 1/32 again
