There are $3,200$ families with $5$ children each. How many of them are expected to have:
a. $3$ boys?
b. at least $4$ boys?
There are $3,200$ families with $5$ children each. How many of them are expected to have:
a. $3$ boys?
b. at least $4$ boys?
Just so you understand better assume that the probability of having a boy is $p$, then that of having a girl is $1-p$. Therefore the probability of having 3 boys is $${{5}\choose{3}} p^3 \cdot (1-p)^{2} $$ Now, multiply this by the number of families to get the answer. Try to do part b on your own. As Jean Marie pointed out in the comments, read up on binomial distribution if you didn't get it.
Events: $A$=having a boy & $B$=having a girl.
Assuming when a family gives birth to a child the P(A) = 1/2 and P(B) = 1/2.50-50 chance
now let's solve (a): let X be a random variable that represent number of boys we first calculate the probability for one family $P(X=3) = $${5}\choose{3}$$*(1/2)^3*(1-1/2)^2 $
now the the answer is $P(X=3) * 3,200$
the same for (b)
let p be probability of having 1 boy child and q be the probability of having 1 girl child
since probability of girls and boys are equal,
therefore
p = 0.5
q = 0.5
P(3 boys and 2 girls) = 5C3 * (p)^3 * (q)^2 = 5C3 * (0.5)^3 * (0.5)^2 = 10/32
P(at least 4 boys) = P(4 boys and 1 girl) + P(5 boys and 0 girl) = 5C4 * (p)^4 * (q)^1 + 5C5 * (p)^5 * (q)^0 = 5C4 * (0.5)^4 * (0.5)^1 + 5C5 * (0.5)^5 * (0.5)^0 = 5/32 + 1/32 = 6/32
no. of families having 3 boys = P(3 boys and 2 girls) * 3200 = 10/32 * 3200 = 1000
no. of families having at least 4 boys = P(at least 4 boys) * 3200 = 6/32 * 3200 = 600