Revision as of 02:16, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Assume that every time you buy a box of Wheaties, you receive one of the pictures of the <math>n</math> players on the New York Yankees. Over a period of time, you buy <math>m \geq n</math> boxes of Wheaties. <ul><li> Use guide:E54e650503#thm 3...")
BBy Bot
Jun 09'24
Exercise
[math]
\newcommand{\NA}{{\rm NA}}
\newcommand{\mat}[1]{{\bf#1}}
\newcommand{\exref}[1]{\ref{##1}}
\newcommand{\secstoprocess}{\all}
\newcommand{\NA}{{\rm NA}}
\newcommand{\mathds}{\mathbb}[/math]
Assume that every time you buy a box of Wheaties, you receive
one of the pictures of the [math]n[/math] players on the New York Yankees. Over a period of time, you buy [math]m \geq n[/math] boxes of Wheaties.
- Use Theorem to show that the probability that you get all [math]n[/math]
pictures is
[[math]] \begin{eqnarray*} 1 &-& {n \choose 1} \left(\frac{n - 1}n\right)^m + {n \choose 2} \left(\frac{n - 2}n\right)^m - \cdots \\ &+& (-1)^{n - 1} {n \choose {n - 1}}\left(\frac 1n \right)^m. \end{eqnarray*} [[/math]]Hint: Let [math]E_k[/math] be the event that you do not get the [math]k[/math]th player's picture.
- Write a computer program to compute this probability. Use this program to find, for given [math]n[/math], the smallest value of [math]m[/math] which will give probability [math]\geq .5[/math] of getting all [math]n[/math] pictures. Consider [math]n = 50[/math], 100, and 150 and show that [math]m = n\log n + n \log 2[/math] is a good estimate for the number of boxes needed. (For a derivation of this estimate, see Feller.[Notes 1])
Notes