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What is the probability that exactly one person is given back the correct hat by a hatcheck person who gives n people their hats back at random?

Short Answer

Expert verified

The probability that exactly one person is given back the correct hat is \(\frac{{n{D_{n - 1}}}}{{n!}} = \frac{{{D_{n - 1}}}}{{(n - 1)!}}\)

Step by step solution

01

In the problem given

Exactly one person is given back the correct hat by a hat check person who gives \(n\) people their hats back at random.

02

The definition and the formula for the given problem

Probability denotes the likelihood of the end result of any random event. The meaning of this term is to test the extent to which any event is probably going to happen. It measures the understanding of the event.

03

Determining the sum in expanded form

Since, exactly one person gets his own hat back, which the remaining \(n - 1\) persons got wrong hate, that is there are \({D_{n - 1}}\) ways to have the remaining hats returned totally incorrectly. Where

\({D_{n - 1}}\)is the number of derangements of \(n - 1\) objects and first person will have \(n\) options to return his hat.

Therefore, the total number of ways to returns a hat among \(n\) persons is the number of different permutations of \(n\) objects is \(n\) ! ways.

Hence, the probability that exactly one person is given back the correct hat is \(\frac{{n{D_{n - 1}}}}{{n!}} = \frac{{{D_{n - 1}}}}{{(n - 1)!}}\).

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Most popular questions from this chapter

Find a closed form for the generating function for the sequence\(\left\{ {{a_n}} \right\}\), where,

a) \({a_n} = 5\) for all\(n = 0,1,2, \ldots \).

b) \({a_n} = {3^n}\)for all\(n = 0,1,2, \ldots \)

c) \({a_n} = 2\)for\(n = 3,4,5, \ldots \)and\({a_0} = {a_1} = {a_2} = 0\).

d) \({a_n} = 2n + 3\)for all\(n = 0,1,2, \ldots \)

e) \({a_n} = \left( {\begin{array}{*{20}{l}}8\\n\end{array}} \right)\)for all\(n = 0,1,2, \ldots \)

f) \({a_n} = \left( {\begin{array}{*{20}{c}}{n + 4}\\n\end{array}} \right)\)for all\(n = 0,1,2, \ldots \)

Express the fast multiplication algorithm in pseudocode.

Prove Theorem 4.

Find the coefficient of \({x^9}\) in the power series of each of these functions.

a) \({\left( {1 + {x^3} + {x^6} + {x^9} + \cdots } \right)^3}\)

b) \({\left( {{x^2} + {x^3} + {x^4} + {x^5} + {x^6} + \cdots } \right)^3}\)

c) \(\left( {{x^3} + {x^5} + {x^6}} \right)\left( {{x^3} + {x^4}} \right)\left( {x + {x^2} + {x^3} + {x^4} + \cdots } \right)\)

d) \(\left( {x + {x^4} + {x^7} + {x^{10}} + \cdots } \right)\left( {{x^2} + {x^4} + {x^6} + {x^8} + } \right.\)\( \cdots )\)

e) \({\left( {1 + x + {x^2}} \right)^3}\)

Find a closed form for the generating function for the sequence\(\left\{ {{a_n}} \right\}\), where

a) \({a_n} = - 1\) for all\(n = 0,1,2, \ldots \).

b) \({a_n} = {2^n}\)for\(n = 1,2,3,4, \ldots \)and\({a_0} = 0\).

c) \({a_n} = n - 1\)for\(n = 0,1,2, \ldots \).

d) \({a_n} = 1/(n + 1)!\)for\(n = 0,1,2, \ldots \)

e) \({a_n} = \left( {\begin{array}{*{20}{l}}n\\2\end{array}} \right)\)for\(n = 0,1,2, \ldots \)

f) \({a_n} = \left( {\begin{array}{*{20}{c}}{10}\\{n + 1}\end{array}} \right)\)for\(n = 0,1,2, \ldots \)

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