Q.4.14 - A First Course in Probability

Question

 A family has n children with probability  $\alpha p^n,n\ge 1$ where $\alpha \le (1-p)/p$

(a) What proportion of families has no children?

 (b) If each child is equally likely to be a boy or a girl (independently of each other), what proportion of families consists of k boys (and any number of girls)? 

Step-by-Step Solution

Answer

The answer of part (a) is 

$P\left(N=0\right)=$$1-\frac{\alpha p}{1-p}$

Part (b) is $P\left(k boys\right)=$$\sum _{n=k}^{\infty }\left(\begin{array}{l}n\\ k\end{array}\right){\left(\frac{1}{2}\right)}^n\alpha p^n$

Step 1:Given Information(Part-a)

The random variable $N$ that denotes the number of children . we know that $N\in N_0$ and that $P(N=n)=\alpha p^n$.for every $n\ge 1$.we are required to find out $P\left(N=0\right)$.

Step 2:Calculation (Part-a)

$P\left(N=0\right)$=$1-P(N\ge 1)$

             =  $1-\sum _{k=1}\alpha p^n$

           =   $1-\frac{\alpha p}{1-p}$

Step 3:Final Answer(Part-a)

The answer is $P\left(N=0\right)=$$1-\frac{\alpha p}{1-p}$

Step 4:Given Information(Part-b)

If we are given that some family has $n$ children, the number of boys in that family has binomial distribution with parameters $n$ and $1/2$.

Step 5:Calculation (Part-b)

$P\left(k boys\right)$=$\sum _{n=k}^{\infty }$ $P\left(k boys/N=n\right) P\left(N=n\right)$

                $=\sum _{n=k}^{\infty }\left(\begin{array}{l}n\\ k\end{array}\right){\left(\frac{1}{2}\right)}^n\alpha p^n$

Step 6:Final Answer (Part-b)

The answer is P(k boys) =$\sum _{n=k}^{\infty }\left(\begin{array}{l}n\\ k\end{array}\right){\left(\frac{1}{2}\right)}^n\alpha p^n$