# child probability

• April 28th 2010, 02:34 PM
drewbear
child 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

• April 28th 2010, 03:05 PM
pickslides
Quote:

Originally Posted by drewbear
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

Lets assume X~Binomial(n=no of trials, p= probability of success)

The binomial theorem states $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

In your case lets call X the probability of having a girl with $n= 20$ and $p = 0.5$

a) find $P(X=0)$

b) exactly 19 boys = exactly 1 girl so find $P(X=1)$

c) "at least 1" means 1 or more so find $P(X\geq 1) = P(X=1)+P(X=2)+\dots+P(X=20)$

Remember $P(A) = 1 - P(A')$

d) "at most 1" means 1 or less so this means 2 or more girls now find $P(X\geq 2) = P(X=2)+P(X=3)+\dots+P(X=20)$