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