child probability

• Apr 28th 2010, 01:34 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

• Apr 28th 2010, 02:05 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 \$\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)\$