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

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- Apr 28th 2010, 01:34 PMdrewbearchild probability
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

- Apr 28th 2010, 02:05 PMpickslides
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 $\displaystyle P(X\geq 2) = P(X=2)+P(X=3)+\dots+P(X=20)$