A family decides to have 20 children. Find the probability of:
a) no girls
b)exactly 19 boys
c)at least 1 girl
d) at most 1 boy
please help and show your work or atleast explain how i can arrive at the answer
Lets assume X~Binomial(n=no of trials, p= probability of success)
The binomial theorem states $\displaystyle P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
In your case lets call X the probability of having a girl with $\displaystyle n= 20 $ and $\displaystyle p = 0.5$
a) find $\displaystyle P(X=0) $
b) exactly 19 boys = exactly 1 girl so find $\displaystyle P(X=1) $
c) "at least 1" means 1 or more so find $\displaystyle P(X\geq 1) = P(X=1)+P(X=2)+\dots+P(X=20) $
Remember $\displaystyle P(A) = 1 - P(A')$
d) "at most 1" means 1 or less so this means 2 or more girls now find $\displaystyle P(X\geq 2) = P(X=2)+P(X=3)+\dots+P(X=20)$