My Math Forum Expected number of children

 June 11th, 2013, 10:19 PM #1 Newbie   Joined: Jul 2008 Posts: 7 Thanks: 0 Expected number of children Hi all, I want to find the answer to the below question ? Q1. In a world, every parent will stop giving birth after they find they have atleast a boy and a girl. Now find what is the expected number of boys and girls in that world? Pr(Boy) = 1/2, Pr(Girl) = 1/2, getting boys and girls are mutually independent. I worked out the answer for this question But i got '2'. But the correct answer is '3'. Please give me the answer with an explanation Thanks in advance Pavan kumar
June 12th, 2013, 07:34 AM   #2
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Re: Expected number of children

Hello, pavankumar.thati!

Quote:
 In a world, every parent will stop giving birth after they find they have at least a boy and a girl. Find the expected number of boys and girls in that world. Pr(Boy) = 1/2, Pr(Girl) = 1/2, getting boys and girls are mutually independent.

$\begin{array}{ccccc}\text{Children}=&\text{Gender}=&\text{Prob.} \\ \\ \\ \hline \\ \\ 2=&BG\text{ or }GB=&2\left(\frac{1}{4}\right) \:=\:\frac{1}{2} \\ \\ \\ 3 &BBG\text{ or }GGB=&2\left(\frac{1}{8}\right) \:=\:\frac{1}{4} \\ \\ \\ 4 &BBBG\text{ or }GGGB=&2\left(\frac{1}{16}\right) \:=\:\frac{1}{8} \\ \\ \\ \vdots &\vdots=$

$\text{Expected number of children: }\:E \;=\;2\left(\frac{1}{2}\right)\,+\,3\left(\frac{1} {4}\right)\,+\,4\left(\frac{1}{8}\right)\,+\,\cdot s$

$\begin{array}{cccc}\text{W\!e have:} && E &=& \frac{2}{2}\,+\,\frac{3}{4}\,+\,\frac{4}{8} \,+\,\frac{5}{16}\,+\,\cdots \\ \\ \\
\text{Multiply by }\frac{1}{2}: && \frac{1}{2}E &=& \;\;\;\;\;\frac{2}{4}\,+\,\frac{3}{8} \,+\,\frac{4}{16}\,+\,\cdots \\ \\ \\
\text{Subtract:} && \frac{1}{2}E &=& 1\,+\,\frac{1}{4}\,+\,\frac{1}{8} \,+\,\frac{1}{16}\,+\,\cdots \\ \\ \\
\text{Multiply by 2:} && E &=& 2\,+\,\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac{1}{8} \,+\,\cdots \end{array}$

$\text{W\!e have: }\:E \;=\;1\,+\,\underbrace{1\,+\,\frac{1}{2}\,+\,\frac {1}{4}\,+\,\frac{1}{8}\,+\,\cdots}_{\text{geometri c series}}$

$\text{The geometric series has first term }a = 1\text{, common ratio }r=\frac{1}{2}
\;\;\;\text{Its sum is: }\,\frac{1}{1\,-\,\frac{1}{2}} \:=\:2$

$\text{Therefore: }\:E \:=\:1\,+\,2 \;\;\;\Rightarrow\;\;\;E\:=\:3$

 June 12th, 2013, 07:44 AM #3 Newbie   Joined: Jul 2008 Posts: 7 Thanks: 0 Re: Expected number of children This is just awesome yet very simple logic. Lots of thanks

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