Question

Question: Use Chebyshev's inequality to show that the probability that more than 10 people get the correct hat back when a hatcheck person returns hats at random does not exceed \(1/100\) no matter how many people check their hats.

(Hint: Use Example 6 and Exercise 43 in Section 7.4.)

Short answer

Answer

\(P(X > 10) \le \frac{1}{{100}}\)

Step-by-step solution

Chebyshev’s Inequality definition

Chebyshev's Inequality: Let \(X\) be a random variable on a sample space \(S\) with probability function \(p\). If \(r\) is a positive real number, then \(p(|X(s) - E(X)| \ge r) \le V(X)/{r^2}\).

Use Chebyshev’s Inequality

If there are total\(n\) no. of people who intend to get their hat back and \(X\) is the random variable that denotes the actual no. of people who do, then it already knows that\(E\left( X \right) = 1,{\rm{ }}V\left( X \right) = 1.\) By Chebyshev's inequality, \(P(|X - E(X)| \ge r) \le \frac{{V(X)}}{{{r^2}}}\). So, the probability that more than 10 people get their correct hat back is

\(P(X > 10) = P((X - E(X)) \ge 10) = P(|X - E(X)| \ge 10) \le \frac{{V(X)}}{{{{10}^2}}} = \frac{1}{{100}}\).

Hence the people check their hat is \(P(X > 10) \le \frac{1}{{100}}\).