Couple has got 5 children , what is the probability of have 3girls and 2 boys ? at least 4 girls? of alternate gender starting with boy ?

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(girls>=4)=P(boys<=1)=P(boys=1)+P(boys=0)
(or P(girls=4)+P(girls=5))

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

A= r2 × 3.14 11×11 ↓ ↓ 121 × 3.14 = 379.94 → 379.9

The correct answer is -20y